AN 


E I T  H  M  E  T I C 


UC-NRLF 


*B    2b5    535 


COLLEGES   AND   SCHOOLS. 


BY  CLAUDIUS  CROZET, 

PRINCIPAL  O?   THE   RICHMOND    A.CADEM\',  LATE  STATE  ENGIKEEK   OF  VIKOINIA, 

AND   ALSO  OF  LOUISIANA  ;     PRESIDENT   OP  JEFFERSON   COLLEGE, 

LOUISIANA,   AND    FORMERLY    PROFESSOR    OF 

KXGI.VEBRING   AT   WEST  POINT. 


RICHMOND,  VA. . 

A  .      MO  B  K  1  S 

1858. 


iif- 


%  (^^ 


IN   MEMORIAM 
FLORIAN  CAJORi 


7; 


AN 


ARITHMETIC 


COLLEGES   AND   SCHOOLS, 


IMPROVED  EDITION. 


BY  CLAUDIUS  CROZET, 

PRINCIPAL   OF   THE   RICHMOND  ACADEMY,  LATE   STATE   ENGINEER   OF  VIRGINIA, 

AND   ALSO    OF  LOUISIANA  ;    PRESIDENT    OF  JEFFERSON   COLLEGE, 

LOUISIANA,   AND  FORMERLY   PROFESSOR   OF 

ENGINEERING  AT   WEST   POINT. 


RICHMOND,   VA.: 
PUBLISHED  BY  A.   MOREIS. 

PHILADELPHIA: 

E.  H.  BUTLER  &  CO. 

1858. 


% 


E-^TBRfiD  ^'(xydtng  fo*ihe  Act  of  Congress,  in  the  year  1848,  by 

CLAUDIUS    CROZET, 

in  the  Clerk's  Office  of  the  District  Court  of  the  United  States  m  and  for  the 
Eastern  District  of  Virginia. 


TABLE  OF  PYTHAGORAS. 


1 

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36 

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32 

36 

40 

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48 

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66 

72 

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28 

35 

42 

49 

56 

63 

70 

77 

84 

8 

16 

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32 

40 

48 

56 

64 

72 

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96 

9 

18 

27 

36 

45 

54 

63 

72 

81 

90 

99 

108 

10 

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30 

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60 

70 

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90 

100 

110 

120 

11 

22 

33 

44 

55 

66 

77 

88 

99 

110 

121 

132 

12 

24 

36 

48 

60 

72 

84 

96 

108 

120 

132 

144 

CAJORI 


PHILADELPHIA  : 

E.  B.  MEARS,  STEREOTYPEa. 

C.  SHERMAN,   PRINTER. 


p 


PKEFACE. 


There  may  be  some  presumption  in  publishing,  with  the 
pretension  of  having  added  some  improvement,  a  work  on  a 
subject  which  has  already  been  treated  by  so  many  authors. 
But  the  late  advances  in  the  mathematics  would  seem  to 
call  for  some  modifications  in  the  manner  of  teaching  their 
first  branch,  so  as  to  make  it  a  more  direct  introduction  to 
higher  subjects,  than  the  Arithmetics  which  are  usually  put 
in  the  hands  of  students.  I  have,  indeed,  frequently  won- 
dered that  a  science  depending  altogether  upon  reasoning  ^.^ 
should  so  generally  be  attempted  tn  he.  tanp;ht  by  practical,  1 
^ rules,  learned  by  rote,  commonly  fQrgotter]^.by  the  pupil  in  a^^      ' 

^few  days,   and   which,   thus  taught,  can  be  correctly  and J 

^readily  applied  only  when,  by  long  usage  and  maturity  of       f 
^mind.   the  arithmetician   has  acquired  a  kind  of  intuitive 
perception  of  the  fundamental  principles  on  which  these 
rules  depend,  and  of  the  train  of  reasoning  by  which  they 
are  obtained. 

I  have  no  hesitation  in  saying  that  if  all  the  time  and  labor 
expended  in  learning "  these  practical  rules  and  acquiring 
readiness  and  skill  in  their  application,  were  bestowed  upon 
the  study  of  the  reasoned  principles  of  the  science,  the  same 
time  and  labor  would  suffice  to  obtain  a  competent  know- 
ledge not  only  of  arithmetic  itself,  but  Hkewise  of  the  two 
other  elementary  branches  of  mathematics;  Algebra  and 
^Geometry. 

Let  a  young  man  begin  arithmetic  at  the  age  when  his 
reasoning  powers  are  sufficiently  matured ;  let  him  then  be 
made  to  investigate  and  comprehend  thoroughly  the  mathe- 
matical principles  jdi  each  operation  as  he  proceeds,  and  I 
venture  to  assert  that  a  few  months  will  suffice  him  to 
become  skilled  in  every  operation  that  can  be  performed  by 
arithmetic  jguided  by  the  unbroken  thread  of  reasoning^ 
he  never  will  torget  IheTuIesT^nd  will  perform  his  calcula- 
tions with  neatness  and  always  by  the  simplest  methods. 

91 82B5 


XV  PREFACE. 

My  experience  has  taught  me  that  it  is  between  the  ages 
of  twelve  and  fifteen^  according  to  the  degree  of  precocity 
of  each  individual,  that  the  study  of  reasoned  arithmetic 
should  begin.  Before  that  period,  a  child  can  only  master 
the  four  plain  rules,  which  he  ought  to  learn  as  soon  as 
practicable;  they  being  for  him  then  a  mere  mechanical 
process.  But  it  is  only  by  long  habit  that  he  can  perform 
with  readiness  these  four  fundamental  operations;  and  there- 
fore he  should  begin  early.  But,  beyond  this  training  of  the 
numerical  faculties  of  the  mind,  I  would  not  torment  a  child 
with  the  abstruseness  of  arithmetic  before  the  above-named 
period. 

In  the  following  pages,  I  have  endeavored  to  avoid  bofh 
the  obscurity  and  confusion  of  purely  practical  arithmetics, 
which  furnish  no  thread  to  guide  the  y*outhful  mind  in  the  per- 
plexing labyrinth  in  which  he  is  frequently  lost,  and  the  other 
extreme  of  abstruseness,  which  some  theoretical  treatises  are* 
obnoxious  to.  My  object  has  been  to  secure  the  knowledge 
of  this  science  in  a  short,  and,  above  all,  in  a  permanent 
way,  and  to  make  the  transition  from  it  to  algebra  simple 
and  natural.  While  I  have  endeavored  to  facilitate  the  pro- 
gress of  the  student,  I  have  not  forgotten  to  make  the  work 
convenient  for  the  teacher,  who  will  find  in  it  numerousP 
carefully  graduated  examples. 

In  these  examples,  I  have  given  the  answers  only  when 
no  part  of  the  operation  is  supplied  by  them.     Answers,  in    ', 
that  case,  are  a  great  assistance  to  the  teacher,  while  they    ^ 
merely  warn  the  pupil  of  his  mistake,  without  doing  away    . 
the  necessity  of  his  going  through  the  details  of  the  exercises. 
'      The  work  is  divided  into  Lessons,  some  of  w^hich  may  heS 
omitted  with  beginners ;  while  others  may  be  only  rapidly 
reviewed  with  more  advanced  pupils,  the  work  being  in- 
tended for  both  classes  of  students. 

This  treatise  will  consist  of  two  parts  : 

The  first,  containing  only  the  elementary  principles  and 
/  fundamental  rules  of  arithmetic. 

I      The  second,  their  combinations  and  applications  to  ques- 
'  tions  of  practical  utility,  and  particularly  the  manner  of  re- 
ducing these  to  numerical  statements. 


TABLE  OF   CONTENTS. 


PART  I. 
CHAPTER    I. 

tESSON  PAOB 

I. — Introduction  and  Definitions 9 

IT. — Spoken  Numeration 11 

III. — Written  Numeration,  or  Notation — ^Decimal  System — Or- 
ders of  Units 13 

IV. — Numeration  continued 15 

V. — Same  subject — Division  of  Numbers  into  Periods          .        .  17 
VI. — How  Numbers  should  be  written  ;  how  they  should  be  read 
— A  Number  is  increased  ten  times  by  moving  its  figures 

one  place  to  the  left 20 

VII.— Roman  Notation 23 


CHAPTER  II. 

CONTAINING   THE    FOUR   EI.EMENTARY   RULES. 

VIII.— Simple  Addition— Its  Proof 24 

IX.— Simple  Subtraction  27 

X. — Same  subject,  continued — Note  on  Subtraction  by  Comple- 
ments— Proof  of  Subtraction 30 

XI. — Simple  Multiplication — Definitions — Nature  of  the  Product 

—Table 34 

XII. — Multiplication,  when  the  multiplier  does  not  exceed  12        .        36 
XIII. — Multiplication  of  Numbers  composed  of  several  Figures — 

Proof 41 

XIV. — Simple  Division — Definitions — Table 46 

XV. — Short  Division — Order  of  the  Quotient — Of  the  Remainder        50 
XVI. — Long  Division — Number  of  Figures  in  the  Quotient — Par- 
tial Dividends 53 

XVII. — Trials  for  each  Figure  in  the  Quotient ;  by  the  common 
method  ;  by  short  division — Case  of  Absent  Orders  in  the 
Quotient — Division  by  multiples  of  ten       ....        56 
XVIII.— Nature  of  the  Quotient- Proof  of  Division— Proof  of  INlul- 

tiplication 61 

XIX.— Quick  Division 64 

1* 


Tl  TABLE   OF    CONTENTS. 

CHAPTER  III. 

CONTAINING    DECIMAL    FRACTIONS. 
LESSON  TPXGZ 

XX. — ^Decimal  Fractions— Supplement  to  Nnntieration— De- 
scending Scale— Notation — Decimal  or  Units' Point  .  67 
XXI.— Reading  of  Decimal  Fractions— Two  ways  of  Reading 
Mixed  Numbers— Writing  of  Decimal  Fractions— Ef- 
fect of  the  Removal  of  the  Units'  Point— Zeros  added 
to  a  Decimal  Fraction  do  not  alter  its  Value     ...        69 

XXn. — Addition  and  Subtraction  of  Decimals 75 

XXIII. — Supplement  to  Multiplication— How  the  Product  is  modi- 
fied by  Increasing  or  Decreasing  its  Factors    ...        77 
XXIV. — Supplement  to  Division— How  the  Quotient  is  modified  by 

multiplying  or  dividing  the  Dividend  and  the  Divisor     .        79 
XXV. — Multiplication  of  Decimals— Remark  on  the   Product  of 

Decimal  Fractions  being  smaller  than  either  Factor       .        81 

XXVI. — ^Division  of  Decimals 86 

XXVII. — On  Decimal  Remainders — Extension  of  the  Quotient — 
Infinite  Decimals — ^Division  of  Decimals  by  Numbers 
ending  with  Os 89 

CHAPTER  IV. 

CONTAINING   VULGAR    FRACTIONS. 

XXVIII.— Nature  of  Vulgar  Fractions— Notation— Definitions  — 
Reading  of  Vulgar  Fractions — Identity  of  Fractions  and 
the  Quotients  of  Division— How  to  Change  a  DecimgJ. 
into  a  Vulgar  Fraction — Fractional  Expressions — Pro- 
per, Improper,  Simple,  Compound,  Complex  Fractions — 

Mixed  Numbers 95 

XXIX. — Fundamental  Propositions — How  a  Fraction  is  Multiplied 

or  Divided  by  a  Whole  Number  .  .  .  .  .  100 
XXX. — Transformation  of  Fractions — Multiplying  or  Dividing 
both  terms  does  not  alter  the  value  of  a  Fraction — What 
is  a  Ratio — Transformation  by  Augmentation  ;  by  Re- 
duction— The  addition  of  the  corresponding  terms  of 
equal  Fractions  gives  a  Fraction  of  the  same  value      .      103 


CHAPTER  V. 

CONTAINING    RULES    RELATIVE    TO    FACTORS    AND   DIVISORS. 

XXXI. — ^Permutations  of  Factors  do  not  alter  the  Product — Con- 
traction in  Multiplication 109 

XXXII. — ^Permutations  of  Divisors  do  not  alter  the  Quotient — Con- 
traction in  Division — Of  Cancelling         ....      112 
XXXIII.— Additional  Definitions— Even  ;    Odd;    Prime  Numbers — 
Common  Divisor — Simplest  Expression  of  Fractions — 
of  Aliquot  Paris  and  Multiples — Power           .        .        .       117 
XXXIV.— Divisibility  of  Numbers  by  Factors,  from  2  to  13      .        .      120 
XXXV.— The  same,  by  Factors,  from  1.3  to  20            ....      124 
XXXVI— General  Rules  of  Divisibility  of  Multiples  and  Factors; 
of  the  whole  and  its  parts — Various  methods  to  ascertain 
the  Divisibility  of  Numbers 126 


TABLE   OF    CONTENTS.  yA 

PART  II. 
CHAPTER  VI. 

CONTAINING   OPERATIONS   IN   VULGAR   FRACTIONS   AND    PROPORTIONS. 
LESSON  PAGE 

XXXVII. — Transformation  of  Fractional  Expressions  and  Numbers, 

or  Reduction 131 

XXXVIII.— Greatest  Common  Divisor 134 

XXXIX. — Transformation  of  Vulgar  into  Decimal  Fractions — ^Re- 
peating or  Periodical  Fractions — Transformation  of  De- 
cimal and  of  Periodical  into  Vulgar  Fractions        .        .      139 
XL. — Reduction  of  Fractions  to  the  sanie  Denominator — Least 

Common  Multiple 144 

XLI. — Reduction  of  Fractions  to  the  Least  Common  Denomina- 
tor  148 

XLTI. — Addition  and  Subtraction  of  Fractions        ....      151 
XLIII. — Multiplication  of  Fractions — How  the  Product  may  be 
smaller  than  either  Factor — Fractions  of  Fractions,  or 

Compound  Fractions 156 

XLIV. — ^Division  of  Fractions 162 

XLV. — Short  Methods  in  Multiplication  and  Division  by  certain 

Numbers — Multiplication  by  Multiples     ....      168 
XLVI. — Proportions  ;    their   Fractional    Character — Definitions — 

Properties — Proportions — Rule  of  Three         .        .        .      174 

CHAPTER  VII. 

CONTAINING   DENOMINATE   NUMBERS. 

XLVIL— Of  Denominate  Numbers — Federal  Money        .        .        .      igo 
XLVIIL— English  Currency— Old  Currency  of  the  States— Tables 
of  Measures  used  in  the  United  States — Remarks  on  the 

same 190 

XLIX. — Transformation  of  Denominate  Numbers           .        .        .      199 
L. — Compound  Numbers ;  their  Transformation  from  one  De- 
nomination to  another 204 

LI. — Transformations  of  both  Vulgar  and  Decimal  Denominate 
Fractions  into  Integers,  and  conversely  of  Integers  into 

Fractions 210 

LII. — Compound  Addition — Proof 215 

LIII. — Compound  Subtraction — Proof     .        .        .        .        .        .      219 

LIV. — Addition  and  Subtraction  of  Denominate  Fractions  .      223 

LV. — Multiplication  of  Compound  Numbers — Remarks  on  the 

Nature  of  the  Multiplicand,  Multiplier,  and  Product      .      225 
LVI. — Compound  Multiplication  by  Aliquot  Parts,  or  Practice, 

when  one  of  the  Factors  is  a  Simple  Number         .        .      231 
LVII. — Compound  Multiplication,  when  both  Factors  are  Com- 
pound   235 

LVIII.— Multiplication  of  Measures  of  Length  and  Duodecimals    .      239 
LIX. — Compound  Division,  when  the  Dividend  and  Divisor  are 

of  the  same  Nature 245 

LX. — Compound  Division,  when  the  Dividend  and  Divisor  are 

of  different  Natures 249 

CHAPTER  VIII. 

CONTAINING   PRACTICAL    QUESTIONS   DEPENDING    ON    PROPORTIONS. 

LXI. — Compound  Proportion,  or  Double  Rule  of  Three — Method 

of  Ratios — Method  of  Units 253 


Vili  TABLE   OF   CONTENTS. 

LESSON  PAGE 

LXII. — Fellowship  or  Partnership           ......  257 

LXIII. — Alligation  Medial ;  Alternate 262 

LXIV. — Percentage— Simple  Interest — Definitions — Rules      .        .  265 
LXV. — Practical  Rules  in  some  cases  of  Simple  Interest,  for  3; 

4  ;  6,  &c.,  per  cent.,  '"or  a  number  of  clays      .        .        .  269 
LXVI. — Simple  Interest,  when  the  year  is  reckoned  at  365  days — 

Interest  on  Sterling  Money 273 

LX VII.— How  to  Calculate  the  Rate,  Time,  Interest        .        .        .276 

LXVIII.— True  Discount 279 

LXIX. — Bank  Discount — Commission — Brokerage — insurance        .  281 

LXX.— Tare  and  Tret— Profit  and  Loss 284 

LXXI. — Exchange  and  Reduction  of  Currencies      ....  286 
LXXII. — Compound  Interest — Table  at  5  and  6  per  o^xit. ;  its  use — 

Rule  to  Calculate  the  Time  for  Doubling  ihe  Capital      .  291 

LXXIII— Compound  Discount 294 

LXXIV. — Annuities — Definitions — Rules— At    Simple    Interest — At 

Compound  Interest 296 

LXXV. — Equations  of  Payments — True  Equated  Time — Common 

Rule— Comparison  between  the  two  Rules     .        .        .  300 
LXXVI. — ^Partial  Payments — General  Practice— Incorrect  and  Irre- 
gular at  Simple  Interest — Remarks  on  Compound  Inte- 
rest— Conclusion 303 


APPENDIX. 


Of  Coins — Currency  of  the  United  States  . 

of  England  .... 

of  France     ..... 

Value  of  Coins  made  Receivable  by  Congress 
Application  of  the  Rules  of  Divisibility  to  Prove  Multiplication  and 
Division  ....... 

Proof  of  Multiplication  by  casting  out  Nines      .  .  , 

Proof  of  Division  by  casting  out  Nines    .... 

Abbreviated  Method  of  Approximation  in  the  Multiplication  of  Deci- 
mals        ........ 

Preparatory  Tables  in  Multiplication  and  Division 


308 
308 


311 
311 
312 

312 
31J 


To  Teachers.^X  few  slight  mistakes  are  purposely  introduced  in  some  of 
the  answers,  in  order  that  the  pupil  may  learn  to  be  sure  of  himself,  and  not 
force  his  results. 


PART   I. 

CHAPTER  I. 

CONTAINING  DEFINITIONS    AND    NUMERATION. 

LESSON  I. 

INTRODUCTION    AND    DEFINITIONS. 

1.  Mathematics  constitute  the  science  which  furnishes 
rules  and  methods  for  measuring  and  calculating  every- 
thing susceptible  of  augmentation  or  diminution. 

2.  Arithmetic  is  the  first  branch  of  mathematics,  and 
confines  itself  altogether  to  numbers :  it  is  the  science  of 
numbers. 

3.  A  Number  is  the  collection  of  several  things  or 
Units,  classed  under  the  same  name. 

4.  A  Unit  is  any  individual  thing :  a  horse,  an  apple, 
one  dollar,  are  different  units. 

The  unit  of  a  number  is  one  of  the  things  it  expresses : 
thus  in  six  cents,  ten  cents,  one  cent  is  the  unit,  six  and 
ten  the  numbers;  in  five  apples,  one  apple  is  the  unit, 
five  is  the  number ;  in  four  hours,  one  hour  is  the  unit, 
four  the  number,  &c. 

Units,  however,  are  seldom  absolutely  such ;  most  fre- 
quently they  have  only  a  relative  character,  and  several 
units  of  a  certain  nature  may  be  collected  so  as  to  form  a 
different  unit  under  another  name. 

Thus  twenty  sheep  may  make  a  unit  under  the  name 
of  one  score;  three  feet  become  a  single  unit  when  called 
one  yard. 

Ttventyfive  units,  called  cents,  make  the  single  unit 
one  quarter. 


10  LESSON  I. 

.One  hundred  cents  are  the  unit  called  one  dollar, 
T^n  dollars  are  fhe  unit,  one  eagle, 
TiveVve  things  make  vi  unit,  as  one  dozen. 
Twelve  dbieh  form  a  new  unit,  as  one  gross,  &c. 
Arithmetic  considers  only  this  numerical  relation  of 
different  units. 

5.  The  particular  unit  to  which  the  value  of  others  is 
referred,  is  called  the  ujiit  of  comparison ;  it  must  be 
carefully  ascertained  before  proceeding  in  any  calcula- 
tion. 

Thus  if  gallons,  quarts,  pints,  or  any  quantity  of  a 
liquid  be  sold,  the  uiiit  of  comparison  will  be  the  gal- 
lon, if  the  liquid  be  sold  by  the  gallon ;  it  will  be  the 
quart,  if  it  sells  by  the  quart,  &c. 

In  the  measurement  of  distances,  the  mile  is  the  unit 
of  comparison. 

In  measuring  timber,  it  is  the  foot  for  the  length,  and 
commonly  the  inch  for  the  breadth,  &c. 

6.  The  unit  of  comparison  is  called  simply  one,  or 
unity,  when  numbers  are  considered  abstractly ;  that  is, 
without  designation  of  the  nature  of  their  units.  Num- 
bers so  considered  are  called  abstract  numbers;  as  when 
we  say,  One,  two,  ten,  one  hundred,  &c. 

7.  When  the  nature  of  the  units  of  a  number  is  desig- 
nated, it  used  to  be  called  a  concrete  number ;  but,  of 
late,  it  has  more  properly  received  the  name  of  denomi- 
nate number;  as  two  dollars,  ten  apples,  one  hundred 
horses. 

8.  The  word  quantity  is  also  frequently  used  instead 
of  number,  to  designate  any  amount  in  a  more  indefinite 
sense.  Thus,  we  say  a  certain  quantity  of  things  ;  this 
quantity  is  greater  than  that,  &c. 

9.  All  the  operations  of  arithmetic  consist  in  disco- 
Tering  and  setting  down  : 

1st.  The  particular  relative  value  of  certain  units. 
2d.  The  quantity  or  number  of  these  units. 


NUMERATION   AND   NOTATION. 


li 


Questio7is. — What  are  mathema;tics  ?  What  is  arithmetic  ? 
What  is  a  number  ?  What  is  a  single  thing  called  ?  In  ten 
cents,  six  apples,  twenty  miles,  &c.,  what  is  the  unit  ?  What  is 
the  number  /  Are  units  absolute  or  relative  ?  Give  examples 
of  relative  units.  What  is  the  unit  of  comparison  ?  Give  ex- 
amples. What  is  an  abstract  number  ?  What  is  a  denominate 
number  ?  Give  examples.  What  is  meant  by  quantity  ?  What 
do  the  operations  of  arithmetic  consist  in  ? 

LESSON  11. 

NUMERATION    AND    NOTATION. 

1.  Numeration  is  the  systematic  expression  of  num- 
bers. 

The  progressive  formation  of  abstract  numbers  consists 
in  joining  one  unit  to  another,  then  one  more  to  this  sum, 
and  thus  continuing  to  add  one  to  each  successive  sum.  In 
this  way,  any  number,  however  large,  may  be  formed. 

The  successive  sums  or  numbers  thus  formed,  must 
each  have  a  particular  name.  All  these  names  compose 
the  spoken  numeration, 

2.  As  there  can  be  no  limit  to  the  formation  of  num- 
bers, it  will  be  conceived  that,  not  only  it  would  be 
impossible  to  recollect  so  many  independent  names,  but 
that  no  use  could  be  made  of  them,  if  these  names  did 
not  convey  to  the  mind  a  ready  and  correct  idea  of  their 
relation  of  value. 

Hence  the  necessity  of  a  system  of  Numeration^  and 
the  importance  of  a  simple  Notation  ;  that  is,  of  a  sim- 
ple method  of  mriting  number s,  consisting  of  a  conve- 
nient nomenclature  and  scale,  by  which  the  relative 
value  of  numbers  may  be  readily  understood  and  applied. 

3.  The  nomenclature  adopted  in  modern  times  forms 
the  decimal  system  of  numeration  (that  is,  of  which  ten  is 
the  base),  chosen  evidently  on  account  of  our  ten  fingers. 

For,  it  was  natural  to  count  successively  on  the  fingers 
One,  two,  three,  four,  five,  six,  seven,  eight,  nine,  ten, 

and  then  to  begin  again  and  go  through  another  ten; 

each  successive  ten  being  recorded  by  a  mark. 

Then  all  the  marks  standing  for  tens,  or  collection  of 

ten  units,  must  be  counted  in  their  turn,  by  means  of  the 


12  LESSON   II. 

same  primary  names ;  so  that  there  would  be  found  suc- 
cessively, 

One  ten,      called  -  -  -  Ten, 

Two  tens,  «  -  -  -  Twenty, 

Three  tens,  "  -  -  -  Thirty. 

Four,  tens,  «  -  -  -  Forty. 

Five  tens,  «  -  -  -  J^ifty- 

Six  tens,  «  -  _  -  Sixty. 

Seven  tens,  *^  -  -  -  Seventy. 

Eight  tens,  «  -  -  .  Eighty. 

Nine  tens,  «  -  -  -  Ninety. 
And,  finally. 

Ten  tens,  '«  -  -  -  O/ie  hundred. 

Another  mark  must  now  be  made  for  a  collection  of 
ten  tens,  the  new  name,  one  hundred,  being  introduced 
for  it ;  and  then  successive  collections  of  ten  tens  may  be 
counted  in  the  same  manner,  under  the  names  of 

One  hundred,  two  hundreds,  &c.,  to  Ten  hundreds, 

counting  by  hundreds,  as  had  been  done  by  tens  and  by 
simple  units. 

By  an  extension  of  the  same  process,  ten  hundreds  re- 
ceive the  new  name  of  one  thousand;  and  every  time 
another  ten  hundreds  is  counted,  an  additional  thousand 
is  recorded,  and  there  is  successively  formed 

One  thousand,  tico  thousands,  &c.,  to  Ten  thousands. 

4.  In  this  way,  each  successive  aggregate  of  ien  is 
counted  as  a  whole,  and  becomes  a  new  unit  of  a  different 
order ;  and  a  scale  is  formed,  in  which  each  new  unit  is 
ten  times  greater  than  the  preceding  one. 

On  this  account. 

Simple  units  are  called  Units  of  the  \st  order* 

Tens,  "  Units  of  the  2d  order. 

Hundreds,  "  Units  of  the  Sd  order. 

Thousands,  '*  Units  of  the  ^th  order. 

Ten  thousands^  '*  Units  of  the  5th  order, 

&c.  &c. 

And  so  on.    Beyond  thousands,  additional  simplifications 


NOTATION.  13 

are  introduced,  which  will  be  better  understood  by  means 
of  the  Notation  or  Written  Numeration. 

N.  B. — When  units  are  mentioned  without  designation  of  any 
particular  order,  simple  units  are  always  understood. 

Questions. — ^What  is  numeration  ?  Spoken  numeration  ?  Writ- 
ten numeration  ?  Notation  ?  Nomenclature  ?  What  is  our  sys- 
tem called  ?  Why  ?  From  what  probable  cause  ?  How  many 
units  form  one  ten  ?  How  many  tens  one  hundred  ?  How  many 
hundreds  one  thousand  ?  What  are  simple  units  called  ?  Tens  ? 
Hundreds?  Thousands?  &c.  How  many  units  of  the  1st  order 
is  a  unit  of  the  2d  order,  of  the  3d,  of  the  4th,  5th,  equal  to  ? 
When  the  word  Unit  is  used  alone,  what  is  understood  ? 

LESSON  III. 

notation. 

1.  The  numbers  in  arithmetic  are  all  expressed  by 
means  of  ten  Arabic  characters^  called  Figures,  which 
were  introduced  into  Europe  by  the  Moors,  about  eight 
or  nine  hundred  years  ago.     They  are  : 


1, 

- 

called 

- 

one. 

2, 

- 

u 

. 

two. 

3, 

- 

cc 

- 

three. 

4, 

- 

u 

- 

four. 

5, 

- 

a 

- 

five. 

6, 

> 

cc 

- 

six. 

•      V, 

- 

a 

- 

seven. 

8, 

- 

a 

- 

eight. 

9, 

- 

cc 

- 

nine. 

0, 

- 

iC 

- 

zero,  cipher,  or  nought. 

N.  B.- 

-Zero 

appears  to  me 

the 

preferable  term.     We  say  the 

thermom 

eter  s 

tands  at  zero  ;  zero 

degree  ;  in  algebra,  a  quantity 

is  made  « 

3qual 

to  zero^  and  not 

ci^Ji 

',er. 

These  figures  were  formerly  all  called  ciphers.  Hence 
the  word  ciphering^  to  express  the  operations  of  arith- 
metic.    Cipher  is  now  used  only  for  0. 

2.  The  first  nine  figures  are  called  signijicant  figures^ 
because  each  signifies  a  certain  number  of  units. 

The  character  0  has  of  itself  no  value ;  it  denotes  only 
the  absence  of  a  thing.     It,  nevertheless,  acts  a  most 


14  LESSON    III. 

important  part  in  the  arithmetical  expression  of  numbers, 
as  will  presently  be  seen. 

3.  Now,  from  what  has  been  said  above  of  spoken 
numeration,  it  will  be  easy  to  conceive  how  any  number 
can  be  written  with  the  nine  significant  figures ;  since  it 
will  be  sufficient  to  set  down  successively  the  number  of 
units,  of  tens,  of  hundreds,  &c.,  it  contains. 

Suppose  that  a  certain  quantity  of  things,  of  eggs,  for 
example,  is  counted,  and  there  are  found 

Six  hundreds^  four  tens^  and  seven  units  (eggs) :  with 
the  figures  we  have  adopted,  it  might  be  written, 

6  hundreds,  4  tens^  and  7  units  (eggs), 
or,  simply,  647, 

by  dispensing  with  the  words  hundreds,  tens,  units;  an 
omission  which  can  produce  no  confusion,  since  these 
names  are  merely  indicative  of  the  order  of  the  units  of 
each  figure,  and  this  order  can  be.  as  readily  understood 
from  the  place  each  figure  occupies : 

The  first  figure  to  the  right  being  7  units  of  the  first  order. 
The  second  is  evidently  4  tens  or  units  of  the  2d  order. 

The  third  "  6  hundreds  or  units  of  the  3d 

order. 

4.  Thus  we  arrive  at  the  fundamental  law  and  simple 
arrangement  of  our  Numeration  :  that 

I.  In  the  Decimal  system  of  Numeration,  the  value  of 
each  Unit  is  made  ten  times  greater  for  each  place  it  is 
removed  to  the  left., 

II.  The  value  of  a  unit  depends  consequently  on  the 
place  it  occupies, 

5.  But  it  may  happen  that  the  number  to  be  written 
may  contain  no  unit  of  a  particular  order.  How,  then, 
if  we  omit  the  names  of  the  different  orders  of  units,  are 
we  to  fix  them  in  the  relative  places  which  determine 
their  values  ? 

By  putting  the  character  0  in  the  place  of  the  absent 
orders  of  units. 


NUMERATION.  l5 

Thus,  in  the  above  example,  if  there  were  no  tens,  the 
number  being  then  6  hundreds  and  7  units^ 
would  be  written  607, 

where  the    6  occupies   still   its  proper  place,  with  the 
assistance  of  0. 

Again,  if  there  were  no  units,  the  number  would  be 
only  6  hundreds^ 

which  would  be  set  down     600. 
That  is,  6  hundreds,  no  tens,  and  no  units, 

6.  In  general,  therefore,  to  write  a  numher  with 
figures,  set  down  the  significant  fi.gures  as  enunciated, 
and  fix  their  orders,  by  filling  up  the  place  of  each 
absent  order  of  units  by  a  zero, 

7.  It  may  be  remarked  that,  in  a  number,  each  signifi- 
cant figure  has  two  kinds  of  values : 

The  first,  its  absolute  value,  which  is  the  intrinsic 
number  of  its  units. 

The  second,  its  relative  value,  which  depends  on  the 
place  it  occupies. 

Questions. — What  characters  are  used  to  express  numbers  ? 
How  are  they  called  ?  By  whom  and  when  were  they  intro- 
duced ?  What  are  the  significant  figures  ?  Why  are  they  so 
called  ?  What  is  the  name  of  the  tenth  character  ?  What  does 
it  denote  ?  How  is  a  number  to  be  expressed  with  the  nine 
significant  figures  ?  What  use  is  made  of  zero  or  cipher  ?  What 
is  the  fundamental  law  of  the  decimal  system  ?  W^hy  is  it  called 
decimal  ?  What  does  the  relative  value  of  a  figure  depend  on  ? 
How  should  a  number  be  set  down  ?  What  is  the  absolute 
value  of  a  figure  ?     What,  its  relative  value  ? 

LESSON  IV. 
1.  According  to  what  has  been  said  in  the  preceding 
Lesson,  of  the  use  of  the  character  0, 


One  ten, 

or 

Ten,       is 

written, 

10 

Two  tens, 

called 

Twenty, 

20 

Three  tens, 

Thirty, 

30 

Four  tens. 

Forty, 

40 

Five  tens, 

Fifty, 

50 

Six  tens, 

Sixty, 

60 

Seven  tens, 

Seventy, 

70 

16  LESSON   IV. 

Eight  tens,       called         Eighty,    is  written       80 

Nine  tens,  "  Ninety^  "  90 

by  placing  0  on  the  right  of  the  significant  figures  1,  2,  3, 

4,  5,  &c.,  which  express  respectively  the  number  of  tens. 

In  the  same  way  will 

One  hundred,  two  hund7*eds,  three  hundreds,  &c., 
be  written  respectively, 

100     ,  200     ,  300     ,         &c. 

One  thousand,     two  thousands,     three  thousands,  &c. 

1000     ,  2000     ,  3000     ,     &c. 

The  hundreds  with  two  ciphers,  to  show  the  absence 
of  both  tens  and  units. 

The  thousands  with  three,  to  show  that  there  are  no 
hundreds,  tens,  nor  units. 

And  the  same  for  any  order,  with  a  corresponding 
number  of  ciphers. 

2.  All  intermediate  numbers  between  10  and  20,  be- 
tween 20  and  30,  &c.,  up  to  One  hundred,  are  expressed, 
as  was  said  before  (III.  3*),  by  setting  down  first  the 
number  of  tens,  and  then  on  its  right  the  number  of 
units.     Thus : 

One  ten  and  one,    cal 

One  ten  and  two,       ' 

One  ten  and  three,    * 

One  ten  and  four,       ' 

One  ten  and  five,       ' 

One  ten  and  six,        ' 

One  ten  and  seven,    * 

One  ten  and  eight,     ^ 

One  ten  and  nine,      ' 

Then  two  tens,  ^ 

Two  tens  and  one,     ' 

Two  tens  and  two,    ' 
&c. 

Any  intermediate  number  between  100  and  200,  200 
and  300,  &;c.,  up  to  one  thousand,  will  be  composed  of 

*  Numbers  between  brackets  are  references  to  other  paragraphs, 
which  the  reader  should  not  omit  to  turn  to.  The  Roman  figures 
indicate  the  Lesson,  and  the  others  the  paragraph  in  it. 


Eleven,       is 

written 

11 

Twelve, 

cc 

12 

Thirteen, 

u 

13 

Fourteen, 

cc 

14. 

Fifteen, 

(C 

15 

Sixteen, 

ii 

16 

Seventeen, 

iC 

17 

Eighteen, 

a 

18 

JSineteen, 

cc 

19 

Twenty, 

C( 

20 

Twenty-one, 

u 

21 

Twenty-two, 

a 

22 

&c. 

&c. 

NUMERATION.  17 

three  figures,  and  written  after  the  same  manner  as  the 
number  647  (III.  3),  by  setting  down  successively  from 
left  to  right  the  figure  of  hundreds^  then  tens,  and  lastly 
units, 

3.  We  can,  therefore,  write  any  number : 
From  one  to  ten,  with  one  figure. 
From  ten  to  one  hundred,  with  two. 

From  one  hundred  to  one  thousand,  with  three,  &;c. 

4.  The  largest  number  with  one  figure  being       9 

"  "  with  two  figures     "       99 

«  «  with  three     "        "     999 

&c.  &c. 

In  general,  the  largest  number  which  can  he  written 
with  any  number  of  figures  is  composed,  of  as  many  9, 
and  one  more  simple  unit  added  to  each  makes  it  respec- 
tively 10,  100,  1000,  &c. ;  that  is,  raises  it  to  a  superior 
unit  of  the  next  order  above  that  number. 

Questions. — How  is  1,  2,  3,  &c.,  tens  written?  How  1,  2,  3, 
4,  &c.,  hundreds?  1,  2,  3,  &c.,  thousands  ?  How  do  you  write 
intermediate  numbers  between  10  and  20,  20  and  30,  &c.  ?  How 
intermediate  numbers  between  exact  hundreds  ?  How  many 
figures  will  express  numbers  under  10?  Between  10  and  one 
hundred  ?  How  many  thence  to  one  thousand  ?  Which  is  the 
largest  in  each  case  ? 

EXERCISES. 

Write  four  tens.  One  hundred  and  twenty.  Fifteen  tens.  4 
single  units,  2  tens,  and  5  hundreds.  6  units  of  the  first  order, 
5  of  the  2d,  9  of  the  3d,  in  one  number.  8  units  of  the  1st,  none 
of  the  2d,  .7  of  the  3d.  Three  hundred  and  forty-five.  Five 
hundred  and  five.  Eight  hundred  and  fifty.  Nine  hundred  and 
ninety-nine ;  write  the  number  formed  by  adding  one  unit  to 
nine  hundred  and  ninety-nine.     Write,  &c.  &c.  &c. 

LESSON  V. 

1.  The  decimal  system  for  numbers,  not  exceeding 
hundreds,  being  well  understood,  its  extension  to  higher 
numbers  presents  no  very  great  difficulty ;  since  we  can 
continue  to  set  units  ten  times  as  great  on  the  left  of  pre- 
ceding ones,  and  thus  reach  to  numbers  of  any  magnitude. 


18  LESSON   V. 

2.  Yet,  if  a  new  name  was  given  to  every  new  order 
of  units,  the  relative  value  of  so  many  different  orders 
of  units  would  not  be  readily  understood. 

On  this  account,  it  being  easy  to  read,  at  once,  three 
figures,  and  to  form  a  ready  idea  of  their  relative  value, 
the  nomenclature  has  been  still  farther  simplified  by 
dividing  numbers  into  sections  of  three  figures,  called 
Periods,  as  in  the  following  example  : 

Trillions*       Billions,        Millio7is.        Thousands,       U7iits, 

304,       120,       571,         054,    64.7 


und; 

ens 

nits 

undi 

sns 

nits 

und] 

sns 

nits 

1=5 

nits 

und] 

?ns 

nits 

^^lip                ^^3p               ^-PS 

^ 

^ 

p           rC  ^:i  s 

The  first  period  being  that  of 

Units. 

The  second,             " 

Thousands. 

The  third,                « 

Millions. 

The  fourth,              « 

Billions. 

The  fifth,                 " 

Trillions. 

Then  Quadrillions,  Quintillions,  Sextillions,  SepdUions, 
Octillions,  Nonallions,  Decallions,  &c.  Beyond  billions, 
however,  numbers  are  too  large  for  any  practical  use,  and 
are  seldom  met  with.    We  count  commonly  then  by  millions. 

3.  Each  period,  it  will  be  seen,  has  its  units,  tens,  and 
hundreds,  and  thus  the  manner  of  expressing  numbers  is 
rendered  very  simple: 

After  having  separated  the  Periods  by  Commas,  each 
Period,  beginning  at  the  left,  is  read  off  successively  as 
a  single  number  of  three  figures,  to  which  the  name  of 
the  period  is  then  given,  and  the  next  one  read  in  its 
turn  to  the  end. 

Thus,  the  number  above  would  read, 

Three  hundred  and,  four  Trillions,  One  hundred,  and 
twenty  Billions,  Five  hundred  and  seventy-one  Millions, 
Fifty-four  Thousands,  Six  hundred  and  forty-seven 
(Units).  The  word  units  is  most  generally  omitted,  it 
being   understood.     When  some   sections   contain   only 


NUMERATION.  f9 

zeros,  their  name  is  omitted,  and  the  significant  figures 
only  enunciated. 

Thus,  4,000,005, 

reads  4  millions  and  5  units,  omitting  the  word  thousands^ 
because  here  that  period  has  no  significant  figure. 

Likewise,      4,005,000, 
reads  4  millions  and  5  thousands,  omitting  the  period  of 
units,  which  contains  no  significant  figure. 

4.  In  order  that  the  mechanism  of  numbers  may  be 
thoroughly  understood,  the  pupil  should  be  exercised  on 
several  examples ;  and,  after  having  divided,  hy  commas, 
the  proposed  number  into  periods,  he  should  be  made  to 
give  to  each  figure  its  proper  name  and  that  of  its  rela^ 
five  valve,  as  follows ;  using  the  same  example  as  above. 

(  7 — units, 
1st  Period,  <  4 — tens, 

(  6 — hundreds. 

r  4 — units  of  thousands, 
2d  Period,  }  5 — tens  of  thousands, 

(  0 — hundreds  of  thousands. 

(  1 — unit  of  millions, 
3d  Period,  <  7 — tens  of  millions, 

(  5 — hundreds  of  millions, 
and  so  on  to  the  highest  period.    This  is  called  numerating, 

5.  He  should  also  be  made  to  designate,  at  once,  any 
of  the  units  of  the  number,  without  going  through  the 
whole  reading,  and  to  repeat  in  order  the  numeration 
scale,  as  follows : 

Units,  Tens,  Hundreds,  Thousands  (or  units  of  thou- 
sands), Tens  of  Thousands,  Hundreds  of  Thousands,  MiU 
lions  (or  units  of  millions),  Tens  of  Millions,  &c. 

Questiojis. — Why  is  a  number  divided  into  periods  ?  How 
many  figures  in  each  period  ?  What  are  their  relative  names  ? 
Name  the  successive  periods.  How  is  a  number  read  ?  Also, 
when  one  or  more  periods  have  no  significant  figure  ?  Is  the  word 
Units  indispensable  after  reading  the  last  period  ?  Read  5,604,325. 
Read  900,807,006  ;  85,000,000,065;  900,000,000,000.  Write  them 
in  words.  Analyze  them.  Repeat  the  numeration  scale.  What 
is  the  name  of  the  5th  order,  of  the  7th,  of  the  10th,  of  the  13th, 
&c.?  In  the  first  number,  what  is  the  relative  value  of  6  ?  of  2  ? 
of  4  ?  of  3  ?     By  what  should  periods  be  divided  ? 


30  LESSON   VI. 


LESSON  VI. 

1.  In  order  to  write  numbers:  Set  down  successively, 
as  they  are  enunciated^  the  three  fgures  ichich  compose 
each  period  ;  taking  care  to  supply  the  place  of  absent 
units  by  0,  and  not  to  omit  any  period ;  the  place  of  those 
which  are  not  enunciated  being  occupied  by  three  Zeros.* 

Exercises, — Write  Five  thousand;  Fifty  thousand;  Five  hundred 
thousand;  Five  millions. 

One  hundred  and  six  thousand  and  nine. 

Three  hundred  and  four  millions,  fifty-four  thousand,  eight 
hundred  and  seven. 

Six  milUons,  one  hundred. 

Nine  billions,  one  hundred  millions  and  forty. 

Five  trillions,  two  hundred  and  four  thousand  and  five. 

Sixty-five  quadrillions  and  seventy  millions. 

Twenty  billions,  six  thousand  and  fifteen. 

N.  B. — It  is  a  good  habit,  which  I  recommend,  never  to  omit 
the  commas  which  divide  periods ;  it  prevents  mistakes,  and 
greatly  facilitates  the  reading  and  arrangement  of  numbers. 

2.  It  may  be  noticed  how  rapidly  the  value  of  the  suc- 
cessive orders  of  units  increases.  Between  the  simple 
unit  One  and  One  ten,  there  is  only  nine  units. 

Between  the  unit  ten,  of  the  second  order,  and  the  unit 
one  hundred  of  the  third  order,  there  is  ninety. 

Between  one  hundred  and  one  thousand,  nine  hundred. 

Between  one  thousand,  the  unit  of  the  fourth  order, 
and  ten  thousand,  that  of  the  fifth  order,  there  is  nine 
thousand. 

*  The  beginner  will  do  well  to  prepare  the  number  he  has  to 
write  as  follows.     Let  it  be,  for  instance, 

Forty  hillionSj  twenty -jive  thousand  and  ten, 
write  successively  in  aline  the  names  of  the  periods  thus: 

Billions,  Millions,  Thousands,  U?iitSi 

with  three  dots  under  each,  separated  by  commas;  then  put  the 
significant  figures  in  their  place,  thus : 

.4.         ,        ...        ,  .25        ,        .1. 

and  then  complete  with  ciphers  for  the  remaining  dots, 

40        ,000        ,025        ,        010 
Ciphers  at  the  beginning  are  omitted  as  useless,  since  they  do  not 
fix  the  relative  place  of  any  figure. 


NUMERATION.  21 

Then  ninety  thousand  between  the  two  nextj  nine 
hundred  thousand  between  the  two  following,  &c. 

So  that  very  few  figures  suffice  to  write  very  large 
numbers. 

3.  It  may  also  be  remarked,  that  the  figure  at  the  head 
of  a  number  has  a  greater  value  than  the  part  of  the 
number  which  follows  it,  since  it  is  composed  of  superior 
units,  each  of  which  is  greater  than  the  largest  number 
that  can  be  formed  with  inferior  units,  which  we  know 
to  be  composed  of  9s  (IV.,  4).  Thus,  in  19  ;  199  ;  1999, 
&c.,  1  ten  is  greater  than  nine  units ;  1  hundred  greater 
than  99;  1  thousand  greater  than  999,  &c.  Consequently, 
it  is  still  greater  than  numbers  composed  of  figures  of 
inferior  value.  Hence,  in  3,975,  2,986,725,  3  has  a 
greater  value  than  975,  and  2  than  986,725,  from  the 
place  they  respectively  occupy. 

Therefore,  of  tivo  numbers^  that  is  the  greater  which 
has  most  figures,  or  whose  first  figure  is  the  largest, 

4.  Not  only  single  units,  but  any  figure  is  loorth  ten 
times  more  for  each  place  it  is  removed  to  the  left ;  and 
consequently,  ten  times  less  for  each  place  it  is  removed 
to  the  right ;  for,  since  the  individual  units  of  which  the 
figure  is  composed  become  ten  times  greater  for  each  re- 
moval to  the  left  (III.,  4),  their  sum  must  likewise  become 
ten  times  greater. 

If,  for  instance,  one  large  cake  weighs  as  much  as  ten 
small  ones,  4  of  the  large  cakes  will  weigh  ten  times  as 
much  as  four  small  ones. 

5.  From  this  remark  we  conclude  that,  to  increase 
(multiply)  a  figure  ten,  one  hundred,  one  thousand  times, 
S^^c,  it  suffices  to  add  respectively,  one,  two,  three  zeros 
to  it,  and,  in  general,  as  many  zeros  as  the  number  of 
times  ten  it  is  to  be  increased. 

Thus,  4  made  ten  times  greater,  is  -  40. 
one  hundred  times,  400. 
one  thousand  times,     4000,  &c. 

6.  Not  only  single  figures,  but  also  whole  numbers, 


22 


LESSON   VI. 


will  be  thus  increased  by  the  addition  of  zeros  on  their 
right. 

Thus,  57,630  is  ten  times  larger  than  5,763  j 
576,300,  one  hundred  times; 
5,763,000,  one  thousand  times,  &c. ; 
since  each  figure,  displaced  one  rank  to  the  left,  is  made 
ten  times   larger  (4),  and  consequently  also  the  whole 
number. 

7.  We  will  close  the  subject  for  the  present  with  an 
important  remark,  viz : 

Any  part  of  a  number  taken  by  itself  forms  a  number 
of  units  of  the  order  of  those  of  its  last  figure.  For  ex- 
ample, in  the  number  6,708,924,  if  we  consider  the  re- 
lative value  of  the  figures  8,  9,  and  2,  we  see  that  2  is  a 
number  of  units  of  the  second  order,  or  tens, 

9  is  composed  of  units  ten  times  as  large,  and  8  of 
units  one  hundred  times  as  large. 

Consequently,  they  stand  in  the  regular  decimal  rela- 
tion in  regard  to  each  other,  and  form  a  number  of  units 
of  the  second  order,  which  may  be  read  892  tens. 

Thus,  again,  in  the  above  number,  we  have, 
67  hundred  thousands, 
670  ten  thousands, 
6708  thousands. 
67089  hundreds,  &c. ; 
the  units  of  the  last  figure  always  giving  their  name  to 
the  number. 

I  have  dwelt  at  some  length  on  Numeration,  because 
experience  has  taught  me  the  importance  of  teaching  it 
thoroughly.  The  student  cannot  devote  too  much  time 
to  this  subject ;  for  it  is  only  when  he  understands  per- 
fectly the  principles  and  mechanism  of  numbers,  that  he 
can  be  expected  to  make  rapid  and  permanent  progress. 

Questions. — How  are  large  numbers  written?  How  is  the 
place  of  absent  periods  supplied  ?  How  many  simple  units  are 
there  between  a  unit  of  the  2d  and  of  the  3d  order  ?  of  the  3d  and 
4th  ?  of  the  4th  and  5th  ?  &c.  Why  is  the  first  figure  of  a  number 
worth  more  than  the  rest  of  the  number  ?  How  much  does  a 
figure  increase  or  diminish  by  change  of  place  in  a  number  ? 
How  is  a  number  increased  ten,  one  hundred,  one  thousand,  ten 


ROMAN    NOTATION. 


23 


thousand  times  ?    How  may  any  section  of  a  number  be  read  by 
itself?    Why  ?    What  name  does  it  take  ? 


LESSON  VII. 
ROMAN  NOTATION. 

1.  Roman  characters  being  frequently  used  for  num- 
bers, the  student  should  be  made  acquainted  with  them. 

The  Romans  expressed  numbers  by  only  seven  capital 
letters  of  the  alphabet,  viz : 


I     for 

-        1, 

C     for 

-        100, 

V     "      -       5,                      D      "      -       500, 

X     "      -     10,                      M     "      -     1000. 

L     "      -     50, 

By  combinations  of  these  letters,  they  formed  all  their 

numbers  after  the  following  manner  : 

I     -    for     1 

XI     -    -    11 

CI    -     -     101 

MI        -     -     1001 

II    -     -     -     2 

XII    -     .     12 

CII-     -     102 

Mil      -    -     1002 

III       -     -     3 

XIII       -     13 

&c.,     to 

&c.,     to 

IlllorlV      4 

XIV  -     -     14 

CC  -     -    200 

MM     -     -     2000 

V    -    -     -     5 

&c.,     to 

CCC     -     300 

MMM       -     3000 

VI-    -     -     6 

XX    -     -    20 

CD  -     -    400 

MV  or  rV      4000 

VII      -     -     7 

XXI  -     -     21 

D     -     -     500 

V    -     -     -     5000 

VIII    -    -     8 

&c.,     to 

DC  -     -     600 

IX.     -    -     9 

XXX      -    30 

DCC     -     700 

VI  -     -     -     6000 

X    -    -    -  10 

XL    -    -     40 

DCCC  -     800 

X     -     -       10,000 

L  -     -     -     /)0 
LX    -     -     60 
LXX       -     70 

CM  -     -     900 
M    -     -  1000 

L     -    -       50,000 
&C.,     to 

LXXX    -     80 

M     -       1,000,000 

XC     -     -     90 

one  million. 

C  -     -     -  100 

In  this  system : 

As  often  as  a  character  is  repeated,  so  many  times 
is  its  value  repealed. 

A  less  character  before  a  greater  is  taken  from  it. 

A  less  character  after  a  greater  is  added  to  it. 

A  bar  over  any  figure  increases  it  one  thousand  fold. 

This  is  evidently  a  very  imperfect  system  of  numera- 
tion. 

Questions. — How  many  characters  were  used  by  the  Romans  ? 
What  were  they  ?     How  are  numbers  expressed  ?     Write  num- 


24  LESSON  vni. 

bers  between  ten  and  twenty ;  between  tbirty  and  forty.  Write 
345;  456;  764,  &c.  What  is  the  value  of  a  character  repeated? 
What  is  the  effect  of  a  less  character  before  a'  greater?  After 
the  same  ?  What  is  a  bar  over  any  character  ?  Write  Three 
millions,  Five  thousand,  five  hundred  and  fifty-five. 


CHAPTER  IL 

CONTAINING   THE    FOUR   FUNDAMENTAL   RULES. 

LESSON  VIII. 

1.  In  every  question  dependent  on  arithmetic,  a  cer- 
tain number,  called  ihe  result,  is  to  be  found. 

2.  The  result  is  ;^enerally  obtained  by  combining  toge- 
ther certain  given  numbers,  or  data^  by  means  of  opera- 
tions which  are  ail  reducible  to  four  elementary  ones, 
namely.  Addition^  Subtraction,  Mjdtiplicafion,  and  Di- 
vision. 

3.  In  these  operations,  several  signs  are  used  as  abbre- 
viations, which  will  be  introduced  and  explained  as  they 
become  necessary. 

The  sign  of  equality  is  =,  and  is  read,  is  equal  to. 
Thus,  one  dollar  =100  cents,  means  and  reads,  one 
dollar  is  equal  to  one  hundred  cents, 

SIMPLE  ADDITION. 

The  sign  is  + ,  called  plus. 

Examples :  2  apples  +  3  apples  =  5  apples. 

2  -f  3  =  5,  read,  2  plus  3  is  equal  to  5. 
Symbols:  2a  +  3a  =  5a.  a-\-b=c. 

Question. — John  has  two  apples,  and  receives  3.  llovf  many 
has  he?    Ans,  2-f  3  =  5. 

4.  Addition  is  an  operation  by  which  several  numbers 
are  united  into  one,  called  their  Sum,  or  total  amount. 


SIMPLE   ADDITION.  25 

There  is  no  difficulty  in  adding  together  single  figures; 
children  at  an  early  day  learn  to  perform  such  additions 
on  their  fingers,  and  to  give  promptly  their  sum.  Pre- 
suming that  this  readiness  has  been  acquired,  and  that  the 
student  will  be  acquainted  with  practical  addition,  before 
a  book  on  arithmetic  is  put  into  his  hands,  I  omit  the 
addition  table. 

5.  Since  a  number  is  a  collection  of  units  of  the  same 
kind,  it  is  clear  that  several  numbers  can  be  added  toge- 
ther so  as  to  form  but  one  number,  only  when  their  units 
are  of  the  same  nature ;  as  when  2  horses  and  3  horses 
are  added  to  make  5  horses;  but  2  horses  and  3  oxen 
could  not  be  united  into  one  number  of  either  horses  or 
oxen. 

Sometimes,  however,  numbers  may  be  presented  with 
different  names,  and  yet  may  be  added  together  if  they 
can  be  brought  under  a  common  denomination. 

Thus,  4  horses,  5  cows,  6  oxen,  may  be  joined  under 
the  same  denomination,  as  15  heads  of  live  stock;  2  lions, 
3  tigers,  and  5  wolves,  as  10  wild  animalsySac, 

6.  Different  units,  therefore,  can  he  added  only  after 
they  have  been  reduced  to  a  common  denomination. 

It  is,  therefore,  always  understood,  in  addition,  that  the 
numbers  are  formed  of  units  of  the  same  kind. 

7.  In  order  to  add  together  several  simple  numbers, 
Write  the  numbers  to  be  added  under  each  other,  so 

that  the  units  of  the  same  order  may  be  exactly  under 
each  other  in  the  same  column.  That  is,  units  under 
units,  tens  under  tens,  hundreds  under  hundreds,  Sfc^ 
and  draw  a  line  beneath  the  last. 

Add  successively,  either  upwards  or  downwards,  the 
figures  of  each  column,  beginning  with  that  of  units, 
and  set  down,  under  the  column  so  added,  the  figure 
which  expresses  the  units  of  its  order. 

If  the  sum  of  the  column  consists  of  more  than  one 
figure,  and  therefore  contains  units  of  the  next  superior 
order,  carry  these  to  the  next  column,  and  add  up  this 
new  column  in  the  same  way. 
3 


26  LESSON    VIII. 

Under  the  last  column  to  the  left,  write  its  entire 
sum, 

3     1  2 

8,  Let  us  take,  for  example,  the  numbers :  5,947 
Be  careful,  if  you  would  avoid  mistakes,  to  859 

place  the  figures  exactly  under  each  other.  3,407 

You  begin  on  the  right,  in  order  that  the  846 

excess  of  each  column  may  be  at  once  added       

to  the  next.  11,059 

Then  you  say  7  and  9  are  16,  and  7  are  23, 
and  6  are  29 ;  and  the  sum  of  the  first  column  being  29 
units,  that  is,  2  tens  and  9  imifs,  you  set  down  the  9 
under  the  column  of  units,  and  carry  the  2  tens  to  the 
next  column  to  which  they  belong. 

The  carrying  to  the  next  column  is  frequently  done 
mentally ;.  but  it  is  a  good  practice  for  beginners,  and  in 
long  additions,  to  set  the  number  carried  at  the  top  of  the 
next  column,  as  shown  in  the  example. 

Passing  on  to  the  second  column,  you  say,  2  carried  and 
4  are  6,  and  5  are  11,  and  0  are  11,  and  4  are  15 ;  that 
is,  15  units  of  the  2d  order,  or  tens,  equal  to  1  hundred 
and  5  tens :  set  down  the  5  under  the  column  of  tens, 
and  carry  1  to  that  of  hundreds. 

Then  you  add  the  third  column,  1  carried  +9  +  8+4 
+  8  =  30  units  of  the  3d  order,  or  hundreds,  which  is 
3  units  of  the  4fth  order,  and  none  of  the  3d;  therefore, 
you  set  down  0  under  the  column  of  hundreds,  and  carry 
3  to  that  of  thousands. 

The  addition  of  the  column  of  thousands  gives  now  11: 
we  set  down  1  and  advance  1  to  the  column  often  thou- 
sands ;  or,  in  other  words,  write  at  once  1 1 ,  as  stated  in 
the  rule. 

PROOF    OF    ADDITION. 

9.  Every  method  of  proving  addition  is,  after  all,  but 
a  repetition  of  the  same  operation.  The  simplest  and 
most  common  way  to  make  this  indispensable  verification, 
and  avoid  the  same  errors,  is  to  reverse  the  order  in 
which  you  have  summed  up ;  if  it  was,  for  example, 
downwards  at  first,  let  it  now  be  upwards. 


SIMPLE    SUBTRACTION.  27 

10.  Sometimes,  also,  one  of  the  numbers  is  cut  off  and 
the  others  added;  after  which  the  single  number  is  also 
added ;  but  this  method  offers  no  advantage. 

When  the  addition  is  very  long,  the  column  had  better 
be  divided  into  several  smaller  ones. 

Questions. — What  is  the  object  of  every  operation  in  arithme- 
tic ?  What  are  the  four  fundamental  rules  ?  What  is  to  be 
found  ?  What  is  given  ?  What  is  the  sign  of  equality  ?  Give 
an  example.  What  is  the  sign  of  addition?  What  is  addition? 
What  is  its  result  called  ?  What  is  the  first  condition,  in  order 
that  several  numbers  may  be  added  ?  How  can  units  of  diiferent 
natures  be  added  ?  Repeat  the  rule  for  addition.  Why  do  you 
begin  on  the  right  ?  If  the  sum  of  a  column  consists  of  one 
figure,  what  is  done  with  it  ?  When  it  consists  of  more,  what  is 
done  ?  What  do  you  carry  ?  Why  ?  Give  examples.  How  is 
addition  proved  ? 

EXAMPLES. 

Examples  for  practice  are  so  easily  set,  and  the  objec- 
tion to  giving  the  results  so  obvious,  that  they  would  be 
of  but  little  use.    They  should  be  given  as  the  following : 

1st.  235 4-697  +  508  + 1016 +  674  + 590=1=3,720. 
2d.  5,709  +  23  +  79,807  +  606  -\-  5  +  709  +  678,753  + 
201+7,900  +  49,722  +  46=  ; 

the  student  being  left  to  arrange  them  in  regular  columns 
himself,  and  setting  down  the  result  after  the  sign  =  in 
the  statement,  which  he  should  copy  and  show,  together 
with  the  details  of  the  operation. 

LESSON  IX. 
SIMPLE  SUBTRACTION. 

1.  Sign  — ,  called  minus. 

Examples :  5  apples  —  2  apples  =  3  apples. 

5 — 2  =  3  read,  5  minus  2  is  equal  to  3. 
Symbols:  5a — 2a=3a.  a  —  b  —  c. 

Qicestion. — John  has  5  apples ;  he  gives  away  2.  How  many 
has  he  left?      A?is.  5— 2  =  3. 

2.  Subtraction  is  an  operation  by  which  the  differ' 
ence  of  two  numbers  is  found. 


28 


LESSON   IX. 


3.  The  result  of  subtraction  is  called  difference^  re- 
mainder^ and  sometimes  excess,  according  to  the  manner 
of  viewing  the  relation  of  the  two  numbers. 

The  smaller  number  is  called  the  subtrahend;  the 
larger,  the  minuend.     These  terms  are  but  little  used. 

4.  Subtraction  is  the  reverse  of  addition;  since,  if  two 
numbers  be  added  together,  one  of  them  taken  away  from 
the  sum,  must  leave  the  other.  For  example  :  John  has 
2  apples,  and  I  give  him  3 )  he  has  then  5.  If,  now,  I 
take  away  from  5  the  3  last  given,  there  remains  evi- 
dently the  2  he  had  at  first. 

5.  In  subtraction,  as  in  addition,  the  numbers  must  be 
composed  of  units  of  the  same  kind,  or  reduced  to  the 
same  denomination,  that  the  operation  may  be  possible 
(VIIL,  6).  No  subtraction  could  be  made,  for  example, 
between  3  horses  and  2  oxen,  except  under  the  common 
name  of  live  stock.  Ten  cents  could  not  be  subtracted 
from  one  dollar,  except  by  considering  the  latter  as  one 
hundred  cents,  &c. 

6.  There  is  no  difficulty  in  subtraction,  when  all  the 
figures  of  the  smaller  number  are  less  than  the  corre- 
sponding ones  of  the  larger. 

Let  it  be  proposed,  for  example,  to  subtract  from  9,587 
if  the  units,  the  tens^  the  hundreds,  &c.,  of  the         345 

second  number  be  respectively  taken  from  the     

units^  tens,  hundreds,  &c.,  of  the  first,  each  part  9,24.2 
will  have  thus  been  subtracted;  and,  conse- 
quently, the  whole  of  the  smaller  number  will  have  been 
taken  from  the  larger.  Commencing,  then,  at  the  right 
hand,  we  •  say  successively,  5  from  7  leaves  2,  4  from  8 
leaves  4,  3  from  5  leaves  2,  and  0  from  9  leaves  9 ;  and, 
setting  down  each  separate  remainder  in  its  proper  place, 
we  find  that  9,242  is  the  whole  remainder ;  so  that,  with 
the  conventional  signs,  9,587—345=9,242. 

7.  But,  in  most  instances,  it  will  happen  that  some  of 
the  figures  of  the  number  to  be  subtracted  are  greater 
than  the  corresponding  ones  of  the  other.  This  gives 
rise  to  the  only  difficulty  in  subtraction. 


SIMPLE    SUBTRACTION.  29 

Let  it  be  proposed,  for  example,  to  find  the 

difference  between  the  numbers      -        -        -  3,756 

and 1,894 


After  having  set  down  the  less  number  under        1,862 
the  greater,  as  in  addition,  so  that  the  units  of 
the  same  order  may  be  under  each  other,  beginning  at  the 
right  hand,  we  say : 

^  from  6  leaves  2,  and  set  down  2  under  the  units. 

Then,  observing  that  9  is  larger  than  5,  add  mentally 
10  to  5,  and  say : 

9  from  15  leaves  6,  and  1  to  carry ;  set  down  6. 

Then  1  carried  and  8  are  9  )  ^  from  17  leaves  8  ;  set 
down  8  and  carry  1. 

Finally,  1  carried  and  1  are  2 ;  2  from  3  leaves  1  : 
set  down  1,  and  the  remainder  is  1,862. 

That  is,  3,756  —  1,894=1,862. 

8.  The  reason  for  carrying  will  be  readily  understood 
by  considering  that,  since  9  cannot  be  subtracted  from  5, 
we  render  the  subtraction  possible  by  adding  to  the  upper 
figure  5  the  convenient  number  10,  which  increases  it  to 
15  ;  from  which  we  subtract  9. 

But  these  ten  units,  which  we  have  added  for  the  con- 
venience of  the  operation,  must  now  be  deducted  in  some 
way.  For  this  purpose,  we  observe  that  they  are  equal 
to  one  unit  of  the  next  higher  order ;  and  then  we  have 
to  subtract,  not  only  the  figure  8  of  the  next  order,  but 
likewise  this  additional  one  ;  that  is,  9. 

The  same  process  is  applied  to  the  new  figure  9,  which  is 
larger  than  7;  we  subtract  from  17,  and  then  deduct  the 
10  that  we  have  mentally  added,  by  carrying  them  as 
1  unit  of  the  higher  order  to  the  figure  of  this  order  in 
the  lower  number. 

Subtraction,  by  carrying  in  this  way,  is  less  liable  to 
errors  than  the  old  method,  formerly  taught,  of  borrow- 
ing;  that  is,  diminishing  the  upper  number  by  one  supe- 
rior unit,  instead  of  increasing  the  lower  one  as  much, 
which  amounts  to  the  same  thing. 
3* 


30  LESSON  X. 

Q2iestio7is. — What  is  subtraction  ?  What  is  its  sign  ?  Give 
an  example.  What  is  its  result  called?  What  is  the  subtrahend ? 
What  is  the  minuend  ?  What  is  the  relation  of  subtraction  to 
addition  ?  What  should  be  the  nature  of  the  units  of  both  num- 
bers ?  How  are  the  numbers  set  down  ?  Where  do  you  begin  the 
operation  ?  How  do  you  subtract  ?  What  is  done  when  the 
lower  figure  is  larger  than  the  upper  ?     Give  a  demonstration. 

LESSON  X. 

1.  From  the  explanations  given  in  the  preceding  les- 
son, we  deduce  the  following  rule : 

1.  Set  down  the  less  number  under  the  greater,*  so 
that  the  units  of  the  same  order  in  each  shall  be  exactly 
under  each  other,  and  draw  a  line  beneath  them. 

II.  Then^  beginning  at  the  right  hand,  subtract  suc- 
cessively each  figure  of  the  lower  number  from  that 
immediately  over  it,  and  set  down  the  remainder, 

III.  When  the  upper  figure  is  the  smaller,  add  in 
your  mind  10  to  it,  and  then  subtract  the  lower  one  ;  set 
down  the  remainder,  and  carry  1  to  the  next  figure  of 
the  lower  number, 

IV.  Continue  the  operation  in  this  way,  until  the  ivhole 
of  the  less  number  has  been  subtracted  ;  if  then  there  are 
more  figures  in  the  larger  number,  bring  them  down  to 
the  left  of  the  remainder. 

2.  The  last  part  of  the  rule  is  exemplified 

in  the  annexed  subtraction,  where,  after  45  has  24,519 

been  subtracted  from  519,  the  remaining  part  45 

24,  of  the  large  number,  since  there  is  no-       • 

thing  to  be  taken  from  it,  is  brought  down  by  24,474 
the  side  of  the  remainder  474,  already  found. 

3.  The  case,  when   the  upper  number  contains   ci- 
phers, must  be  noticed.      Take  this  ex- 
ample:         200,010,025 

Here  we  say,  8  from  15  leaves  7;  1  car-  30,008 

ried  from  2  leaves  1 ;  then  0  from  0,  twice,     • • 

leavesO;  3 from  11  leaves 8; then, carrying     199,980,017 

*  This  is  customary ;  but  it  will  he  well  to  exercise  the  pupil  to 
Bublract  in  any  relative  position  of  the  numbers. 


SIMPLE   SUBTRACTION.  3l 

1,  though  there  is  no  remaining  figure  in  the  lower  num- 
ber, we  continue  to  sa}^,  1  from  10,  three  times,  leaving  as 
many  9s  as  there  are  zeros  above  j  and,  finally,  1  from  2 
leaves  1.     Hence, 

When  there  are  zeros  above  and  below  (the  absence 
of  figures  in  the  less  number  being  equivalent  to  zeros), 
if  you  do  not  carry  1,  set  down  at  once  in  the  result  as 
many  Os  ;  and^  if  you  cari'y^  as  many  9s  as  there  are  Os 
•m  the  upper  number  ;  taking  care^  in  this  case^  to  de* 
duct  \  froiYh  the  figure  tohich  precedes  the  Os. 

4.  It  frequently  happens  that  there  is  no  remainder 
under  the  left-hand  figures  of  the  two  numbers :  in  that 
case,  do  not  set  down  ciphers  in  the  remainder. 

For  example :  -  -  -  24,506 

24,497 

9 

Simple  dots,  to  show  that  no  column  has  been  omitted, 
are  sufiicient.  Zeros  at  the  beginning  of  a  number  are 
useless ;  the  practice  is  awkward,  and  may,  in  some  in- 
stances, cause  mistakes  to  be  made. 

Note. — There  is  a  manner  of  subtracting  which  beginners 
cannot  well  understand,  but  for  which  I  cannot  find  a  more  suit- 
able place.     It  is  by  means  of  the  Arithmetical  complement. 

The  Arithmetical  complement  of  a  number  is  its  difference  from 
1,  with  as  many  zeros  as  the  number  itself  has  figures. 

Thus  the  arithmetical  complement  of  -  -  -  67,814 
is     - 32,186 


their  sum  being  ----,-_       100,000 

The  Arithmetical    complement  is    obtained    by  stcbtr acting  the 

figure  of  units  from  10,  and  all  the  oilier s  from.  9. 

Suppose,  now,  that  you  had  to  subtract  from  -         98,653 

•  67,814 


30,839 
Add  to  the  upper  number  successively  each  figure  of  the  arith- 
metical coTnplemeyit ,  carrying  as  iLsual^  and  at  the  end  strike  02ct 
the  superior  luiit  carried. 
Thus  we  would  say, 

instead  of  4  from  13, 6  +  3=9 

instead  of  2  from  5, 8  +  5  =  13 

instead  of  8  from  16,  -        -        -     1  carried,  +  1  +  6  =   8 

Then, 2  +  8=10 

and  ----..     1  carried,+ 3  + 9  ~i-^ 


32  LESSON  X. 

The  last  1  is  struck  out,  and  the  answer  is  as  above,  30,839. 
The  reason  of  this  is  that, 

98,653— 67,814  =  98,653  +  32,186—  100,000; 

which,  for  those  who  do  not  understand  algebra,  may  be  explained 
by  the  following  illustration  : — A  man  owes  me  36  dollars,  and  I 
owe  him  97  ;  difference,  61. 

I  give  him  his  97  dollars,  but  he  cannot  pay  back  my  36  dol- 
lars otherwise  than  by  giving  me  a  100  dollar  note ;  for  which  I 
make  the  change  64,  which  is  the  arithmetical  complement  of  36, 
and  then  the  account  stands  thus  as  received  by  him  : 

97  4-  64  —  100  =  161  —  100  =  61. 

This  method  is  expeditious  when  several  numbers  are  added, 
from  which  one  is  to  be  subtracted.     Thus,      -        -      1,217,245 

-f  612,940 

4-  726,783 

—  1,675,321 


=  ^0  881,647 

It  may  be  also  used  advantageously  in  subtracting  several 
numbers  at  once  from  the  sum  of  several  others ;  only,  in  this 
case,  strike  out  the  last  superior  unit  carried,  for  each  number. 
Example:         -        -        -        768,403 

+  297,546 

4-  829,427  :  that  is,  you  add, 

—  562,973  -    -    -    137,027 

—  428,465  -    -    -    571,535 

—  356,893  -    -    -    643,107 

—  263,563  -    -    -    736,437 
and  cut  off  4. 

=^283,482, 

PROOF    OF    SUBTRACTION. 

Add  the  remainder  to  the  less  number;  the  sum  must 
he  equal  to  the  greater  number,  if  the  work  is  right 
(IX.,  4). 

EXAMPLE. 

From        -        -        5,386,427 
take       -        -        4,258,792 


Remainder,  1,127,635 


Proof,  -        5,386,427 


SIMPLE   SUBTRACTION.  33 

Questions. — Repeat  the  rule  of  subtraction.  What  is  done  if 
there  are  more  figures  in  the  greater  number  than  in  the  lower  ? 
When  both  contain  ciphers  I  When  the  figures  on  the  left  leave 
no  remainder?     How  do  you  prove  subtraction?     Why? 

EXAMPLES    FOR   PRACTICE. 
1.  24,607,801  —  15,608,819=  8,998,982. 


7.  1,000  —  7  = 

8.  4,000  —  1  = 

9.  30,000  —  15=. 

10.  10,000,000  —  10  = 

11.  45,001  —  11== 


2.  1,234,567  —  702,973  = 

3.  7,020,914  —  2,766,809  = 

4.  95,000  —  1,006  = 

5.  4,073,050,062  —  2,803,767,- 
086  = 

6.  20,004,001,003  —  8,405,- 
128,605  = 

12.  From  fifty  thousand  and  sixty-three  take  one  thousand  and 
sixty-five. 

13.  From  two  hundred  millions  and  three  thousand  subtract  one 
million  and  nine. 

14.  The  flood  happened  about  the  year  of  the  world  1656,  and 
the  birth  of  Christ  4004.     How  long  was  it  before  Christ? 

How  long  before  the  present  year  ? 

15.  Gunpowder  was  invented  in  1330 ;  printing  in  1441.  How 
long  did  the  first  precede  the  second? 

16.  America  was  discovered  in  1492.     ITow  long  ago  is  it  ? 

17.  The  mariner's  compass  was  invented  in  1302.  How  long 
did  it  precede  the  discovery  of  America? 

18.  General  Washington  died  in  1799.  How  long  has  he  been 
dead? 

19.  The  first  settlement  in  the  United  States  was  made  by  the 
English,  at  Jamestown,  Virginia,  in  1607.  How  long  after  the 
discovery  of  America  ? 

How  long  before  the  Declaration  of  Independence,  in  1776  ? 

20.  I  have  lent  a  man  2,000  dollars;  he  has  paid  back  at  seve- 
ral times  250,  314,457,  and  510  dollars.  How  much  does  he  owe 
yet? 

21.  A  merchant  buys  600  barrels  of  flour,  for  3,600  dollars, 
and  sells  450  barrels  for  3,150,  how  many  barrels  are  left  ?  And 
how  much  more  must  he  receive  to  get  back  his  money  ? 

22.  A  merchant  buys  20,750  yards  from  one  person,  16,184 
from  another,  3,066  from  a  third;  he  sells  successively   1,200 

C 


1,609, 

and  creditor. 

346 

6,567, 

-   12,612 

12,676, 

9,709 

983, 

28 

13,910, 

5,600 

56, 

4,315 

34f  LESSON   XI. 

yards,  7,965  yards,  10,704,  and  671.  How  many  yards  has  he 
left? 

23.  I  have  a  revenue  from  rents  of  1,570  dollars ;  from  stock, 
1,600.  I  spend  for  house-rent,  450;  for  fuel,  360;  for  provisions, 
875;  for  the  doctor,  75;  servants'  hire,  320;  taxes,  87;  in- 
surance, 107;  clothing,  550.  How  much  have  I  left  at  the  end 
of  the  year  ? 

,  24.  A  man  pays  350  dollars  for  100  sheep,  60  for  a  pair  of 
oxen,  200  for  two  horses,  450  for  a  carriage.  He  gives  in  return 
200  bushels  of  wheat,  for  185  dollars;  two  cows,  worth  45;  50 
barrels  of  corn,  worth  275.     How  much  has  he  to  pay,  besides  ? 

25.  A  merchant  finds  himself,  in  his  account 

With  A,     in  debt, 

«  B,  « 

«  C,  « 

«  D,         " 

«  E,         " 

«  F,         " 
Does  he  owe  more  than  is  due  to  him,  and  how  much  ? 

The  last  question  is  one  of  those  which  afford  means  of  verifi- 
cation by  varying  the  operation.  For,  it  is  clear  that  we  may 
by  subtraction  find  the  balances  (differences)  in  each  individual 
account,  add  them  on  their  respective  sides,  and  then  subtract. 

Or  else  add  the  debts  and  credits,  and  subtract ;  both  of  which 
methods  must  produce  the  same  answer.  For,  it  is  the  same 
thing  whether  he  pay  all,  and  receive  from  all  together,  or  settle 
with  each  separately. 

In  all  the  above  questions,  after  having  ascertained  what  the 
nature  of  the  units  of  the  answer  will  be,  the  numbers,  in  the 
operation,  are  considered  as  abstract  numbers. 


LESSON  XL 
SIMPLE  MULTIPLICATION. 

1.  Sign  X]  read,  into  or  multiplied  by;  sometimes, 
simply  hy. 

Examples:  2  X  4*  =  8,  read,  2  multiplied  by  4,  are 
equal  to  8. 

Symbols:  a  X  b  =  c, 

2.  Question. — John  gives  6  apples  to  each  one  of  his  4  brothers. 
How  many  apples  has  he  given  in  all  ?     Answer. — 24  apples. 

Or,  1  yard  of  cloth  costs  6  dollars.  How  much  will  4  yards 
cost  ?     Ans2ver.-^2i:  dollars. 


SIMPLE   MULTIPLICATION.  35 

Such  questions  as  these  give  rise  to  multiplication.  In 
both  cases,  6  is  to  be  repeated  4  times ;  which  might  be 
done  by  adding  6  +  6  -}-  6  -f  6 =24.  But  it  is  more  simple 
to  say,  4  times  6  are  24.     This  is  Multiplication. 

3.  As  regards  the  manner  of  performing  the  operation, 
we  have  this 

FIRST    DEFINITION. 

Multiplication  is  a  short  method  of  Addition^  hy  I 
which  a  number  is  repeated  as  many  times  as  there  are 
units  in  another. 

In  the  above  case,  for  instance,  4  is  repeated  6  times, 
and  we  have,  4x6  =  24. 

4.  But  when,  from  the  nature  of  a  question,  we  wish 
to  ascertain  the  character  of  the  operation,  we  may  use 
this 

SECOND    DEFINITION. 

Multiplication  is  an  operation  by  which  the  amount 
for  One  unit  being  given,  that  for  a  certain  number  of 
the  same  units  is  obtained. 

In  the  above  1st  Question,  for  example,  we  see  that 
One  brother  has  6  apples ;  the  4  brothers,  consequently, 
must  have  4  times  6  apples. 

In  the  2d,  One  yard  is  worth  6  dollars  j  4  yards  must 
be  worth  4x6. 

5.  The  number  to  be  repeated  is  called  the  Multipli- 
cand.    It  is  the  amount  for  One  unit  of  the  other.. 

The  number  by  which  it  is  multiplied,  is  the  Multi- 
plier. One  of  the  units  of  the  multiplier  is  equivalent 
to  the  whole  multiplicand. 

The  result  of  the  operation  is  called  the  Product. 
Both   the    Multiplicand  and  the    Multiplier  are    also 
called  Factors  of  the  Product  j  because  it  is  formed  by 
multiplying  them  together. 

In  the  above  example,  6  is  the  Multiplicand ; 
4  "  the  Multiplier  5 
24  "the  Product; 
6  and  4  are  the  two  Factors  of  the 
Product, 


36 


LESSON   II. 


6.  Sometimes,  also,  the  product  is  called  a  Mnltijple 
of  one  of  its  factors,  to  express  that  it  contains  it  an 
exact  number  of  times,  and  can  therefore  be  obtained  from 
it  by  multiplication. 

Thus,  24  is  a  multiple  of  6,  because  4  times  6  make 
24 )  but  27  is  not. 

7.  Since  it  is  the  multiplicand  which  forms  the  pro- 
duct, by  being  repeated  a  number  of  times,  it  follows 
that  the  product  is  of  the  same  nature  as  the  multipli^ 
cand;  and,  since  the  number  of  times  it  is  repeated  is 
marked  by  the  units  of  the  multiplier : 

1.  That  when  the  multiplier  is  one,  the  product  is 
equal  to  the  multiplicand, 

2.  That  when  the  multiplier  is  more  than  unity,  the 
product  is  larger  than  the  multiplicand, 

3.  And  by  analogy,  that  when  the  multiplier  is  less 
than  unity,  the  product  is  smaller  than  the  multiplicand. 
This  occurs  in  fractions,  as  will  be  seen  hereafter. 

8.  We  will  consider  three  cases  in  multiplication  : 
The  first,  when  neither  of  the  two  factors  exceeds  12. 
The  second,  when  one  of  them  has  several  figures,  and 

the  other  does  not  exceed  12. 

The  third,  when  both  factors  are  composed  of  several 
figures. 

CASE  I. 

9.  This  case  requires  only  a  ready  knowledge  of  the 
following  table  of  multiplication,  which  must  be  com- 
mitted to  memory : 


1  time 

0 

is 

0 

Twice 

0 

are 

0 

3  times 

0 

are 

0 

1  time 

1 

is 

1 

Twice 

1 

are 

2 

3  times 

1 

are 

3 

1  time 

2 

is 

2 

Twice 

2 

are 

4 

3  times 

2 

are 

6 

1  time 

3 

is 

3 

Twice 

3 

are 

6 

3  times 

3 

are 

9 

1  time 

4 

is 

4 

Twice 

4 

are 

8 

3  times 

4 

are 

12 

1  time 

5 

is 

5 

Twice 

5 

are 

10 

3  times 

5 

are 

15 

1  time 

6 

is 

6 

Twice 

6 

are 

12 

3  times 

6 

are 

18 

1  time 

7 

is 

7 

Twice 

7 

are 

14 

3  times 

7 

are 

21 

1  time 

8 

is 

8 

Twice 

8 

are 

16 

3  times 

8 

are 

24 

1  time 

9 

is 

9 

Twice 

9 

are 

18 

3  times 

9 

are 

27 

1  time 

10 

is 

10 

Twice 

10 

are 

20 

3  times 

10 

are 

30 

1  time 

11 

is 

11 

Twice 

11 

are 

22 

3  times 

11 

are 

33 

1  time 

12 

is 

12 

Twice 

12 

are 

24 

3  times 

12 

are 

36 

SIMPLE  MULTIPLICATION. 


37 


4  times 

0 

are 

0 

7  times 

0 

are 

0 

10  times 

0 

are 

0 

4  times 

1 

are 

4 

7  times 

1 

are 

7 

10  times 

1 

are 

10 

4  times 

2 

are 

8 

7  times 

2 

are 

14 

10  times 

2 

are 

20 

4  times 

3 

are 

12 

7  times 

3 

are 

21 

10  times 

3 

are 

30 

4  times 

4 

are 

16 

7  times 

4 

are 

28 

10  times 

4 

are 

40 

4  times 

5 

are 

20 

7  times 

5 

are 

35 

10  times 

5 

are 

50 

4  times 

6 

are 

24 

7  times 

6 

are 

42 

10  times 

6 

are 

60 

4  times 

7 

are 

28 

7  times 

7 

are 

49 

10  times 

7 

are 

70 

4  times 

8 

are 

32 

7  times 

8 

are 

56 

10  times 

8 

are 

80 

4  times 

9 

are 

36 

7  times 

9 

are 

63 

10  times 

9 

are 

90 

4  times 

10 

are 

40 

7  times 

10 

a>e 

70 

10  times  10 

are 

100 

4  times 

11 

are 

44 

7  times 

11 

kre 

77 

10  times  11 

are 

110 

4  times 

12 

are 

48 

7  times 

12 

are 

84 

10  times  12 

are 

120 

5  times 

0 

are 

0 

8  times 

0 

are 

0 

11  times 

0 

are 

0 

5  times 

1 

are 

5 

8  times 

1 

are 

8 

11  times 

1 

are 

11 

5  times 

2 

are 

10 

8  times 

2 

are, 

10 

11  times 

2 

are 

22 

5  times 

3 

are 

15 

8  times 

3 

are 

24 

11  times 

3 

are 

33 

5  times 

4 

are 

20 

8  times 

4 

are 

32 

11  times 

4 

are 

44 

5  times 

5 

are 

25 

8  times 

5 

are 

40 

11  times 

5 

are 

55 

5  times 

6 

are 

30 

8  times 

6 

are 

48 

11  times 

6 

are 

66 

5  times 

7 

are 

35 

8  times 

7 

are 

56 

11  times 

7 

are 

77 

5  times 

8 

are 

40 

8  times 

8 

are 

64 

11  times 

8 

are 

88 

5  times 

9 

are 

45 

8  times 

9 

are 

72 

11  times 

9 

are 

99 

5  times 

10 

are 

50 

8  times 

10 

are 

80 

11  times 

10 

are" 

110 

5  times 

11 

are 

55 

8  times 

11 

are 

88 

11  times  11 

are 

121 

5  times 

12 

are 

60 

8  times 

12 

are 

96 

11  times  12 

are 

132 

6  times 

0 

are 

0 

9  times 

0 

are 

0 

12  times 

0 

are 

0 

6  times 

1 

are 

6 

9  times 

1 

are 

9 

12  times 

1 

are 

12 

6  times 

2 

are 

12 

9  times 

2 

are 

18 

12  times 

2 

are 

24 

6  times 

3 

are 

18 

9  times 

3 

are 

27 

12  times 

3 

are 

36 

6  times 

4 

are 

24 

9  times 

4 

are 

36 

12  times 

4 

are 

48 

6  times 

5 

are 

30 

9  times 

5 

are 

45 

12  times 

5 

are 

60 

6  times 

6 

are 

36 

9  times 

6 

are 

54 

12  times 

6 

are 

72 

6  times 

7 

are 

42 

9  times 

7 

are 

63 

12  times 

7 

are 

84 

6  times 

8 

are 

48 

9  times 

8 

are 

72 

12  times 

8 

are 

96 

6  times 

9 

are 

54 

9  times 

9 

are 

81 

12  times 

9 

are 

108 

6  times 

10 

are 

60 

9  times 

10 

are 

90 

12  times 

10 

are 

120 

6  times 

11 

are 

66 

9  times 

11 

are 

99 

12  times  11 

are 

132 

6  times 

12 

are 

72 

9  times 

12 

are 

108 

12  times 

12 

are 

144 

Questions. — What  is  the  sign  of  multiplication  ?  Give  an  ex- 
ample. What  questions  give  rise  to  multiplication?  What  is 
multiplication?  What  is  the  multiplicand?  The  multiplier? 
What  is  the  result  called  ?  What  are  factors  ?  To  vt'hich  does 
the  given  unit  belong  ?  What  is  it  equal  to  ?  Of  what  nature  is 
the  product?    When   is   it  equal  to  the  multiplicand?     When 


38  LESSON   XI. 

larger?    When  smaller?     How  much  is   7X8j   6x7 j   9X5; 
12X6;  2X6X7;  3x2x11,  &c.? 

EXERCISES. 

I.  How  much  are  12  pounds  of  sugar,  at  6  cents  a  pound? 
•  2.  How  much  are  8  pounds  of  coffee,  at  12  cents  a  pound? 

3.  How  much  are  7  yards  of  cloth,  at  5  dollars  a  yard  ? 

4.  How  much  are  9  pencils,  at  2  cents  a  pencil  ? 

5.  How  much  are  11  pairs  of  shoes,  at  3  dollars  a  pair? 

6.  How  much  are  one  dozen  pairs  of  stockings,  at  4  shillings 
a  pair  ? 

7.  How  much  are  10  chickens,  at  9  cents  apiece  ? 

8.  How  much  are  9  bushels  of  wheat,  at  6  shillings  a  bushel  ? 

9.  There  are  5  bushels  of  wheat  in  a  barrel.  How  many  in 
9  barrels  ? 

10.  At  9  dollars  a  month,  what  will  be  my  wages  in  a  year  ? 

II.  9  rows  of  trees;  9  trees  in  a  row.  How  many  trees  in 
all? 

12.  A  man  hires  a  cart,  horse,  and  driver;  the  cart  and  horse 
at  12  shillings,  the  man  at  10  shillings  a  day.  What  has  he  to 
pay  ior  11  days? 

13.  Bought  11  yards,  at  9  cents;  6  yards,  at  8  cents;  7  yards, 
at  10  cents.     What  is  to  pay? 

14.  Bought  9  pieces  of  cloth,  containing  11  yards  in  a  piece; 
6  pieces,  containing  12  yards;  7  pieces,  containing  8  yards.  How 
many  yards  in  all  ? 

15.  A  man  buys  11  sheep,  at  3  dollars,  and  sells  them  at  4. 
What  does  he  gain  ?  What  would  he  lose,  if  he  sold  them  for 
25  dollars  ? 

In  all  these  questions,  the  pupil  will  easily  discover  that  a  cer- 
tain number  is  given  as  equivalent  to  the  unit  of  another  number; 
and  that,  after  having  determined  from  the  data  the  character 
of  the  product,  the  multiplication  proceeds  as  with  abstract 
numbers. 

Let  the  pupils  answer,  in  each  case,  these 

Questions. — What  is  the  unit  of  comparison?  What  is  the 
nature  of  the  multiplicand?  The  multiplier?  What  is  the 
nature  of  the  product? 

Such  examples  may  be  multiplied  with  beginners,  and  omitted, 
as  well  as  this  and  other  simple  lessons,  with  more  advanced 
pupils. 


SIMPLE   MULTIPLICATION.  SS 

LESSON  XII. 
CASE  II. 

1.  When  one  of  the  factors  does  not  exceed  12 : 

1.  Set  down  the  small  number  as  multiplier  under  the 
large  number,  and  draw  a  line  beneath, 

II.  Multiply  each  figure  of  the  multiplicand  by  the 
multiplier^  setting  down  and  carrying  as  in  addition, 

FIRST   EXAMPLE. 

7  houses  were  sold  at  5,869  dollars  each.  What  is 
the  whole  cost  7 

Here,  5,869  is  the  multiplicand ;  and  the  product  must 
be  in  dollars  (XI.,  7).  Having  ascertained  this,  we  pro- 
ceed to  repeat  5,869,  as  an  abstract  number,  7  times. 

2.  The  result  might  be  obtained  by  the 
successive  addition  of  each  figure  of  the 
above  number  7  times,  with  proper  carry- 
ing. But  we  shorten  the  operation  by 
multiplication,  and  say : 

7  times  9  are  63  ;  set  down  3  and  carry  6. 

7  times  6  are  42,  and  6  carried,  are  48 ; 
set  down  8  and  carry  4. 

7  times  8  are  56,  and  4  carried,  are  60  j  set  down  0 
and  carry  6. 

Finally,  7  times  5  are  35,  and  6  carried,  are  41  j  which 
set  down. 

3.  The  arrangement  of  the  successive  units  set  down 
in  the  product,  is  the  consequence  of  the  principle 
(VIII.,  5)  that  the  repetition  of  units  of  any  kind  or 
order,  must  give  units  of  the  same  hind  or  order, 

4.  After  having  ascertained,  from  the  nature  of  the 
question,  what  the  nature  of  the  product  will  be,  it  has 
been  said  that  the  multiplication  is  made  as  if  with  abstract 
numbers. 

5.  Furthermore,  to  simplify  the  operation,  we  have 
recommended  to  consider  the  smaller  number  as  the  mul- 


OPERATION. 

5,869 
7 

41,083 


1 

1 

1 

1 

1 

1 

1 

1 

1 

1 

1 

1 

1 

1 

1 

1 

1 

1 

1 

1 

1 

1 

1 

1 

40  LESSON   XII. 

tiplier.  This  we  are  justified  in  doing,  from  the  princi 
pie  that,  in  multiplying  two  abstract  numbers  together, 
the  product  is  the  same,  whether  the  first  he  multiplied 
by  the  second,  or  the  second  by  the  first ;  or,  in  other 
words,  it  is  indifferent  which  of  the  factors  is  made  the 
multiplier. 

This  fact  may  have  been  remarked  in  the  Multiplica- 
tion Table  j  and  may  be  further  understood,  in  general, 
as  follows : 

Suppose  that  We  place  6  units  in 
a  horizontal  row,  and  form  four 
such  rows;  We  shall  have  four 
times  6  units. 

But  the  simple  inspection  of  the 
diagram  will  show  that  the  same  might  have  been  done 
by  forming  6  vertical  rows,  of  4  units  each,  which  is  6 
times  4  units ;  equal,  therefore,  to  4  times  6. 
Hf:nce,  4x6  =  6x4. 

It  is  clear  that  this  illustration  would  apply  to  any  two 
numbers,  and,  consequently,  that  the  principle  is  general. 

6.  The  same  may  also  be  explained  as  follows : 

I  give  4  apples  to  each  of  6  boys ;  it  is  in  all  6  times 
4  apples. 

But,  instead  of  this,  I  might  give  each  boy  1  apple 
first,  making  in  all  6  apples,  and  repeat  it  four  times, 
which  would  be  four  times  6  apples;  and,  since  each  boy 
will  then  have  4  apples,  as  before  j  it  is  evident  that  6 
times  4  =  4  times  6. 

SECOND   EXAMPLE. 

What  is  the  cost  of  7,006  yards  of  cloth,  at  9  dollars 
per  yard  ? 

Here  it  is  9  dollars  whicji  is  to  repeated  7,006  times, 
and  which  is  the  multiplicand.  But,  availing  ourselves  of 
the  preceding  remarks,  and  considering  the  two  factors  as 
abstract  numbers,  we  make  9  the  multiplier,  and  say : 

9  times  6  are  54 ;  set  down  4  and     1     ^  qqq 
carry  5.  '     g 

9  times  0  are  0,  and  5  are  5.  

9  times  0  are  0.  dollars. 

9  times  7  are  63.  ^^^054 


SIMPLE   MULTIPLICATION. 


41 


Q2ir,stio7is , — How  do  you  multiply  by  one  figure?  What  do 
units  of  any  order,  multiplied  by  simple  units,  give?  Of  two 
numbers,  which  is  the  multiplicand  ?  Will  it  alter  the  result  to 
take  it  for  the  multiplier?     Prove  this. 


EXAMPLES   FOR   PRACTICE. 


I      6,007X5  = 

7,974x8  = 

30,457X9  = 

60,812X6  = 

908,004x7  = 

287,815X9  = 


123,456,789x2  = 
123,45S,789X3  = 
123,456,789X4  = 
123,456,789x5  = 
123,456,789X6  = 
123,456,789x7  = 


123,456,789  X  8  = 
123,456,789  X  9  = 
123,456,789X10  = 
123,456,789X11  = 
123,456,789X12  = 
67,400,000 X   7  = 


LESSON  XIII. 

CASD  III. 

1.  To  multiply  by  a  number  composed  of  several 
figures. 

1.  Choose  of  the  two  factors  that  which  appears  most 
convenient  as  multiplier  (XII.,  5) ;  set  it  doton  under  the 
other,  considered  as  multiplicand ,  and  draw  a  line 
beneath. 

II.  Form  successively  th$  product  of  the  midtiplicavd 
by  each  of  the  significant  figures  of  the  multiplier,  and 
write  the  individual  products  under  each  other,  beneath 
the  line,  observing  to  place  the  first  right-hand  figure  of 
each  product  in  the  column  directly  under  its  midtiplier, 
and  the  other  figures  on  the  left,  under  preceding  nuni' 
bers,  so  as  to  form  regular  columns,  as  in  addition. 

III.  Add  up  the  partial  products,  as  they  are  ar- 
ranged  ;  their  sum  will  be  ths  total  product  required. 

2.  Let  it  be  proposed,  for  example,  to  multiply 


In  order  to  understand  the  se* 
cond  part  of  the  rule,  we  must 
consider  that  we  have  to  repeat 
the  multiplicand,  657,  as  many 
times  as  there  are  units  in  the 
multiplier,  849  ;  that  is,  800  times 
+  40  times  4"  9  times. 

In  the  first  place,  we  know 
4* 


by 


657 
849 


product  by  9,  5,913 
"  by  40,  26,280 
"    by  800,  525,600 

total  product,     557,793 


42 


LESSON   XIII. 


already  how  to  multiply  by  9  simple  units,  and  obtain  the 
product,  5,913. 

3.  But  how  is  the  multiplication  by  40  to  be  effected'^ 
Evidently  by  repeating  the  multiplicand  40  times;  that 
is,  ten  times  4. 

Hence,  after  having  multiplied  by  4 ;  that  is,  repeated 
the  multiplicand  4  times,  and  found  2,628,  we  must  take 
this  product  10  times.  The  result  will  be  40  times  the 
multiplicand. 

But  we  have  seen,  in  Numeration  (VL,  4,  5,  6),  that 
to  increase  a  number  10  times,  we  have  only  to  add  one  0 
to  it,  by  which  the  units  become  tens. 

Hence,  we  multiply  now  the  product  2,628  by  ten,  by 
adding  a  0,  which  makes  it  26,280,  and  advances  the 
figure  8  of  simple  units  to  the  column  of  tens,  and  thus 
places  it  under  its  multiplier  4,  to  the  order  of  which  it 
is  raised  by  the  zero. 

In  like  manner,  the  multiplication  by  800  is  made  by 
adding  two  Os  to  the  product  of  the  multiplicand  by  8  j 
since  this  product  must  be  one  hundred  times  that  by  8, 
and  consequently  of  the  order  of  hundreds. 

4.  From  this  explaAiation  We  conclude  that :  The  mvU 
tiplication  of  a  nt^filvr  of  simple  units  hy  iinits  of  any 
order,  gives  in  the  product  units  of  the  same  order  :  that 
is,  the  multiplicalior;  of  units  by  tens,  gives  tens;  by  hun- 
dreds, gives  hundreds  J  &c. 

The  same  may  also  be  understood  by  remarking  that, 
to  multiply  a  number  of  units  by  hundreds,  is  the  same 
thing  as  multiplying  the  hundreds  by  that  number  (XII.,  5) , 
and,  consequently,  must  produce  hundreds. 

In  practice,  the  Os^,  which  pioperly  belong  to  each  par- 
tial product,  are  omitteil,  as  unneces- 
sary, since  the  order  of  the  first  figure 
is  sufficiently  indic^vted  by  its  place  in 
the  column  of  its  order.  The  opera- 
tion is  commonly  set  do\vn  thus : 

But  beginners  will  do  well  to  retain 
these  Os  until  they  can  multiply  cor- 
rectly without  them. 


657 

849 


5913 
2628 
5256 

557793 


SIMPLE   MULTIPLICATION.  43 

5.  It  is  customary  to  multiply  in  regular  order,  begin- 
ning with  the  units.  This  is  more  natural  and  conve- 
nient, but  not  indispensable;  and  any  arrangement  might 
be  used,  provided  the  figures  of  the  partial  products  be 
placed  in  their  proper  column,  we  might,  for  example, 
multiply  -  -  -  657 

by        849 

Thus,  product  by  4  tens,  -  26280 
"  by  8  hundreds,  525600 
«      by  9  units,       -         5913 

Total  product,  -   557,793 

6.  The  above  rule  applies  to  all  cases  of  multiplication. 
It  may  not  be  amiss,  however,  to  consider  particularly 
the  case,  when  there  are  ciphers  at  the  right  hand  of  one 
or  both  of  the  factors. 

The  rule  applies  as  readily  to  this  case ;  for  then,  also, 
the  place  of  the  units  here  occupied  by  a  0,  must  be  ad- 
vanced by  the  proper  number  of  Os,  to  the  order  of  the 
first  significant  figure  you  multiply  by.  Let  it  be  pro- 
posed, for  instance,  to  multiply 

47000 
by       1900 


42300000 
47000 


The  result  is  89,300,000. 

That  is,  the  product  423000  of  the  multiplicand  by  9 
units  of  the  3d  order  is  advanced  to  the  place  of  hundreds 
by  two  additional  zeros ;  and  that  by  1  follows. 

It  will  be  readily  observed  that  to  the  ciphers  of  the 
multiplicand  there  is  thus  added  in  the  product  those  of 
the  multiplier ;  so  that  there  is  in  the  product  a  number 
of  ciphers  equal  to  those  of  both  the  multiplicand  and 
multiplier.     Hence, 

When  there  are  ciphers  at  the  end  of  one  or  both  of 
the  factors^  rnultiply  as  if  they  were  not  there,  and  then 
add  to  this  product  as  many  ciphers  as  there  are  in  both 
factors. 


4>4f  LESSON   XIII. 

7.  Remark. — When,  as  in  the  last  example,  one  of  the  figures 
you  have  to  multiply  by  is  1,  do  not  go  through  the  process  of 
multiplication,  but  set  down  at  once  the  whole  multiplicand, 
since  any  number  multiplied  by  1,  is  evidently  that  number 
itself. 

This  remark  is  inserted  here,  because  I  have  very  frequently 
seen,  even  advanced  pupils,  go  through  an  awkvi^ard  regular  multi- 
plication by  1 ! 

PROOF    OF    MULTIPLICATION. 

8.  The  only  proof  of  multiplication  which  can  be 
understood  at  present,  besides  going  over  the  same  work, 
is  to  make  the  multiplicand  the  multiplier ;  and  if  the 
products  are  alike,  the  operation  may  be  presumed  to  he 
correct. 

Questions, — Give  the  general  rule  of  multiplication.  Explain 
why  the  right-hand  figure  of  each  partial  product  is  to  be  set 
down  under  its  multiplier.  When  you  multiply  simple  units  by 
a  figure  of  any  order,  what  is  the  order  of  the  product  i  Why 
are  the  additional  Os  of  partial  products  omitted  in  practice  ? 
How  do  you  multiply  when  there  are  ciphers  at  the  end  of  one 
or  both  of  the  factors  ?  Why  ?  How  do  you  multiply  by  1?  By 
10?  By  100?  By  1000?  What  is  the  product  of  units  by  tens? 
By  hundreds  ?  Of  tens  by  tens  ?  By  thousands  ?  Of  hundreds 
by  hundreds  ?  Of  thousands  by  thousands  ?  Of  hundreds  by  tens 
of  thousands?  Of  hundreds  by  tens  ?  Of  thousands  by  hundreds 
of  thousands  ?  Of  thousands  by  millions  ?  &c.  How  do  you 
prove  multiplication  ? 

EXERCISES. 

1.  36X25=900.  2.  9,664X16  =  154,624. 

3.  365X38  =  13,870.  4.  41,364X35  =  1,447,740. 

5.  421X48=20,208.  6.  34,293X74=2,537,682. 

7.  895X23=20,585.  8.  91,738x81=7,430,778. 

9.  5,296X29  =  153,584.  10.  576,784x64  =  36,914,176. 

11.  2,564X47  =  120,508.  12.  718,328x96  =  68,959,488. 

13.  540,042X23  =  12,420,966.     14.  9,378,964x42  =  393,916,488. 

15.  1,345,894X49=65,948,806. 

16.  12,345,679  X  27  =  333,333,333. 

17.  46,123,101x72=3,320,863,272. 

18.  14,565,869X51  =  742,859,319. 


SIMPLE   MULTIPLICATION.  45 

19.  5,004,000x60=  20.  5,8.')lx  657  =3,844,107. 

21.  2,843,200X80=  22.  37,864X209=7,913,576. 

23.  814X951=774,114.  24.  40,058x342  =  13,699,836. 

25.  594x437=259,578.  26.527,527x285=150,345,195. 

27.  1,345X108  =  145,260.  28.  749,643X695  =  521,001,885. 

29.  1,055,054  X  570  =  601,380,780. 

30.  40,200,050X320  =  1,286,401,600. 

31.  87,468X5,847=  511,425,396. 

32.  5,401X3,004  =  16,224,604. 

33.  34,960,078X8,405  =  293,839,455,590. 

34.  529,600X2,900  = 

35.  12,321x12,321  =  151,807,041. 

36.  523,972 x  15,276  =  8,004,196,272. 

37.  1,055,070X31,456=33,188,281,920. 

38.  500,407  X  870,497  =  435,602,792,279. 

39.  6,795,634X918,546=6,242,102,428,164. 
40.40,070,065x38,015,732  =  15,232,906,283,422,580. 

41.  2,708,630,425X206,008,604  =  558,001,172,606,176,700. 

These  examples  may  also  be  used  for  Pivision,  by  giving  to 
the  pupil  the  product  and  one  factor. 

They  may  also  serve  for  exercises  with  ciphers,  by  adding 
some  at  the  end  of  the  factors. 

For  additional  questions,  see  Division  and  Appendix. 

1.  There  are  360  degrees  round  the  earth;  each  degree  mea- 
sures 121,519  yards.  What  is  the  circumference  of  the  earth,  in 
yards  ?  A7is.  43,746,840  yards. 

2.  A  man  bought  a  farm  of  235  acres,  at  56  dollars  an  acre ;  a 
second,  of  320,  at  60  dollars;  a  third,  of  628,  at  27  dollars;  and 
a  fourth,  of  497,  at  39.     How  much  did  he  pay  for  the  whole? 

Ans.  68,699  dollars, 

3.  A  locomotive  runs,  every  day,  16  miles  an  hour,  for  twelve 
hours.     How  many  miles  has  it  travelled  in  three  years? 

A?is,  210,240  miles. 

4.  238  chests  of  tea,  of  45  pounds  each,  at  2  dollars  per  pound. 

98  bags  of  coffee,  at  23  dollars  per  bag^r 
56  firkins  of  butter,  at  19  dollars  per  firkin. 
129  hogsheads  of  molasses,  at  27  dollars. 
^*kXt  is  the  whole  amount  ?  Ans.  28,221  dollars. 


46  •  '  LESSON   XIV. 

5.  21  pieces  of  cloth,  of  36  yards  each,  at  7  dollars  a  yard. 
49         "  «  67         "  9  " 

37         «  «  29         "  5  « 

58         "  «  73         "  13  " 

69         «  "  59         "  6  " 

How  many  pieces,  how  many  yards,  and  what  is  the  whole  cost  ? 
A71S,  23i  pieces  J  13,417  yards,  119,672  dollars. 

In  the  first  question,  the  amount  of  yards  in  one  degree  is  given, 
and  that  for  360  to  be  found ;  in  the  second,  the  cost  of  o?ie  acre, 
and  that  of  many  required ;  in  the  third,  the  run  in  one  hotir  is 
given,  and  that  in  the  number  of  hours  contained  in  three  years, 
required. 

In  one  word,  each  question  gives  the  amount  for  o7ie,  and 
requires  it  for  many ;  showing  the  operations  to  be  multiplica- 
tions. 

LESSON  XIV. 
SIMPLE  DIVISION. 

1.  Signs :  2^,  or  24  :  4  =  65  read,  24  divided  by  4,  is 
equal  to  6. 

Symbols ;  -J  =  c,  a  :  b  =  c  ;  a  divided  by  b,  is  equal 
to  c. 

Qjiestions. — John  has  24  apples,  which  he  divides  equally 
among  his  4  brothers.     How  many  does  he  give  to  each  ? 

Ansiuer,  6  apples. 

Or,  4  yards  of  cloth  have  cost  24  dollars.  How  much  is  it  a 
yard  ?  Answer,  6  dollars, 

2.  Questions  like  these  give  rise  to  Division,  which, 
it  will  be  perceived,  is  the  inverse  rule  of  Multiplication. 
In  that  operation,  two  numbers  are  given,  and  the  aggre- 
gate of  their  product  required )  in  this,  the  aggregate  is 
known  as  well  as  one  of  the  numbers,  and  the  other  is 
to  be  found.     In  other  words,  we  see  that 

FIRST   DEFINITION. 

3.  Division  is  an  operation  by  which  the  product  and 
one  of  its  factors  being  Jinown,  the  other  factor  is  to  be 
found. 

In  questions  like  the  above,  for  example,  we  must 


SIMPLE   DIVISION.  47 

ascertain  how  many  times  the  small  number  4  is  contained 
in  the  larger,  24. 

This  might  be  done  by  successive  subtractions  of  4,  by 
which  it  would  be  found  that  4  goes  6  times  in  24  j  and 
that,  consequently,  6  is  the  answer. 

But  it  is  shorter  to  say  at  once.  In  24,  how  many  times 
4  ?  6  times,  is  the  ready  answer.  This  method  changes 
subtraction  into  Division, 

As  regards,  therefore,  the  mere  mechanism  of  the  ope- 
ration, our  first  impression  of  its  character  is  that 

SECOND   DEFINITION. 

4.  Division  may  he  considered  as  a  short  method  of 
Subtraction^  by  which  it  is  found  how  many  times  a 
number  is  contained  in  another. 

In  the  above  case,  4  is  contained  6  times  in  24.  Hence, 
we  have  2_i  =  6,  for  the  answer. 

But  when,  from  the  nature  of  the  question,  we  wish 
to  ascertain  what  operation  is  to  be  performed,  it  is  more 
readily  understood  by  reference  to  the  First  Definition. 

For,  whenever  an  aggregate  amount  is  given,  as  well  as 
one  of  the  numbers  which  have  produced  it,  we  know, 
at  once,  that  it  is  by  division  that  the  other  is  to  be  found. 

We  know,  for  example,  that  4  brothers  have  24  apples 
among  them,  and  we  wish  to  find  the  share  of  each. 

Or,  we  know  that  24  dollars  is  the  total  cost  of  4  yards, 
and  wish  to  find  the  cost  of  one ;  the  answer  in  both 
cases,  repeated  4  times,  must  produce  24. 

In  these  examples,  4  and  6  are  the  factors  of  24. 

5.  The  number  to  be  divided  is  the  Dividend.  It 
corresponds  to  the  Product  of  multiplication. 

The  number  w^e  divide  by,  the  Divisor. 
The  result  is  called,  the  Quotient.    Both  are  the  factors 
of  the  Dividend. 

6.  Consequently,  the  dividend  contains  the  divisor  as 
many  times  as  there  are  units  in  the  quotient, 

24  contains  4  as  many  times  as  there  are  units  in  6. 


48 


LESSON   XIV. 


Hence,  1st.  When  the  divisor  is  one,  the  quotient  is  a 
number  equal  to  the  dividend, 

2d.  When  the  divisor  is  greater  than  unity,  the  quO' 
tient  is  a  nvmher  smaller  than  the  dividend, 

3d.  When  the  divisor  is  smaller  than  unity,  the  quo- 
tient is  a  number  larger  than  the  dividend, 

7.  Sometimes  the  divisor  does  not  go  an  exact  number 
of  times  in  the  dividend,  and  there  is  left  a  number  smaller 
than  the  divisor.     This  is  called  the  Remainder, 

If,  for  example,  we  divide  27  by  4,  we  get  6  to  24, 
and  3  over. 
mainder. 


27  is  the  dividend,  4  the  divisor,  3  the  re- 


8.  We  will  consider  two  cases  in  division :  the  first, 
Short  Division,  wht^n  the  divisor  does  not  exceed  12. 

The  second,  Long  Division,  when  the  divisor  exceeds 
12. 

They  both  require  a  ready  knowledge  of  the  following 


DIVISION  TABLE. 


2 

in 

2 

once 

3 

in 

4 

twice 

2 

in 

6 

3  times 

2 

in 

8 

4  times 

2 

in 

10 

5  Times 

2 

in 

12 

6  times 

2 

in 

14 

7  times 

2 

in 

16 

8  times 

2 

in 

18 

9  times 

5 

in 

5 

once 

5 

in 

10 

twice 

5 

in 

15 

3  times 

5 

in 

20 

4  times 

5 

in 

25 

5  times 

5 

in 

30 

6  times 

5 

in 

35 

7  times 

5 

in 

40 

8  times 

5 

in 

45 

9  times 

6     in 
6     in 


in  3  once 

in  6  twice 

in  9  3  times 

in  12  4  times 

in  15  5  times 

3     in  18  6  times 

3     in  21  7  times 

3     in  24  8  times 

3     in  27  9  times 


in 
in 


b     once 
12     twice 


6     in     18     3  times 
6     in     24    4  times 

30 

36 


5  times 

6  times 


6  in  42  7  times 
6  in  48  8  times 
6    in    54    9  times 


4 

n 

4 

once 

4 

;n 

8 

twice 

4 

n 

12 

3  times 

4 

n 

16 

4  times 

4 

m 

20 

5  times 

4 

in 

24 

6  times 

4 

m 

28 

7  times 

4 

in 

32 

8  times 

4 

m 

36 

9  times 

7 

n 

7 

once 

7 

n 

14 

twice 

7 

m 

21 

3  times 

7 

in 

28 

4  times 

7 

n 

35 

5  times 

7 

in 

42 

6  times 

7 

m 

49 

7  times 

7 

in 

56 

8  times 

7 

in 

63 

9  times 

SIiMPLE    DIVISION. 


4>9 


8 

in 

8 

once 

9 

in 

9 

once 

10 

in 

10 

once 

8 

in 

16 

twice 

9 

in 

IS 

twice 

10 

in 

20 

twice 

8 

in 

24 

3  times 

9 

in 

27 

3  times 

10 

in 

30 

3  times 

8 

in 

32 

4  times 

9 

in 

36 

4  times 

10 

in 

40 

4  times 

8 

in 

40 

5  times 

9 

in 

45 

5  times 

10 

in 

50 

5  times 

8 

in 

48 

6  times 

9 

in 

54 

6  times 

10 

in 

60 

6  times 

8 

in 

56 

7  times 

9 

in 

63 

7  times 

10 

in 

70 

7  times 

8 

in 

64 

8  times 

9 

in 

72 

S  times 

10 

in 

SO 

8  times 

8 

in 

72 

9  times 

9 

12 

in 

81 

9  times 

10 

in 

90 

9  times 

in 

11 

once 

in 

12 

once 

Questioiis, 

in 

22 

twice 

12 

in 

24 

twice 

30_ 

44 

in 

33 

3  times 

12 

in 

36 

3  times 

5    - 

-J 

1  1 

in 

44 

4  times 

12 

in 

48 

4  times 

72  _ 

- 

5  5 

in 

55 

5  times 

12 

in 

60 

5  times 

Sd  - 

in 

66 

6  times 

12 

in 

72 

6  times 

T  - 

TT  — 

in 

77 

7  times 

12 

in 

84 

7  times 

6  3_ 

y  — 

- , 

GO 

T2  — 

8  4 

in 

88 

8  times 

12 

in 

96 

8  times 

96  _ 

* 

in 

99 

9  times 

12 

in 

108 

9  times 

12- 

v= 

25- 
5    - 

\^  = 

48  _ 
6-  - 

=  ? 

T2  — 

V- 

4 
42 

G 

32  _- 

4 

Questions. — What  is  the  sign  of  division?  Give  an  example. 
What  questions  lead  to  division  i  What  is  division  ?  First 
definition  ?  Second  definition  ?  What  is  the  dividend  ?  The 
divisor?  The  quotient?  The  remainder?  Give  examples. 
How  many  times  does  the  dividend  contain  the  divisor  ?  What 
is  the  quotient  when  the  divisor  is  1  i  More  than  1  ?  Less 
than  1  ?  How  many  cases  in  division  ?  In  ...  .  how  many 
times  .  .  .  .  ?     What  is  ...  .  divided  by  ....  ? 


EXERCISES. 

1.  A  man  travels  12  miles  a  day;  how  long  will  it  take  him 
to  travel  144  miles  ?  The  aggregate  amount  is  given,  and  or,  ^ 
of  its  factors ;  the  other  to  be  found.  Hence  the  operation  is 
Division. 

2.  A  man  gets  56  dollars  for  8  sheep. "  How  much  a  sheep  ? 
The  amount  for  several  is  given,  that  for  1  to  be  found  :  Division, 

3.  How  many  oranges  can  be  bought  for  84  cents,  at  6  cents 
apiece?  The  product  and  one  factor  known;  the  other  to  be 
found  :  Division, 

4.  A  hatter  has  to  pack  72  hats,  and  can  put  9  in  a  box.  How 
many  boxes  will  he  want  ?     The  whole  amount,  72,  is  known  j  9 

5  D 


50  LESSON   XV. 

must  be  one  of  its   factors;  the  other  factor  is  the  answer: 
Divisio7i, 

5.  Four  persons  make  up   132  years  with  their  joint  ages. 
What  is  the  average  of  their  ages  ? 

6.  7  men  bought  25  horses  for  2000  dollars,  and  sold  them  for 
1923.     How  much  did  each  lose  ? 

7.  John  has  47  apples,  and  gives  8  to  each  of  5  children.    How 
many  has  he  left  ? 

8.  Such  questions  as  these  are  to  be  asked : 

In  84,  how  many  times  5  ?  and  how  much  over  ? 

9.  How  much  is  87  divided  by  9  ?  and  how  much  over  ? 
Varying  the  numbers  in  both. 

10.  A  man  paid  48  dollars  for  8  sheep,  and  99  for  9  cows. 
How  much  more  did  he  pay  for  each  cow  than  for  one  sheep  ? 


LESSON  XV. 
SHORT  DIVISION. 

1.  Question. — To  divide  4326  dollars  equally  he'- 
tween  7  persons. 

In  order  to  obtain  a  correct  idea  of  the  operation,  we 
must  consider  this  number  as  composed  of  the  following 
parcels : 

4  of  one  thousand  dollars  each. 
+  3  of  one  hundred. 
+  2  of  ten. 
H-  6  of  one. 
And  now  we  have  to  ascertain  how  many  of  each  parcel 
or  order  the  share  of  each  person  will  contain. 

Beginning  with  the  thousands,  as  there  are  but  four  of 
them  for  7  persons,  they  cannot  have  any  entire  thousand 
in  their  individual  shares. 

We  must,  therefore,  pass  on  to  the  next  inferior  order, 
and  change  the  higher  units  into  smaller  ones  by  consi- 
dering that  4  thousands  =  40  hundreds.  These  we  join 
to  the  3  in  the  given  number,  which  makes  in  all  43 
hundreds.  These  divided  by  7,  give  6  hundreds  for  each 
share,  and  1  hundred  over. 

This  hundred  also  must  be  divided :  for  this  purpose, 
it  is  changed  into  10  tens,  which,  joined  to  the  two  tens 


SHORT    DIVISION.  51 

in  the  given  number,  make  12  tensj  and  these  divided 
by  7,  give  to  each  1  ten,  and  there  remain  5  tens  over. 

The  remainder,  5  tens,  joined  to  the  6  units  of  the  num- 
ber, makes  56  :  and  56  divided  by  7,  gives  exactly  8  sim- 
ple units  to  each.  So  that  the  share  of  each  person  is 
composed  of  6  hundreds, 

1  ten, 
and   8  units. 
That  is,  collectively,  618  dollars. 

2.  The  operation  is  performed  in  the  following  man- 
ner: 

Place  the  divisor  on  the  right^  of  the 
dividend.     Separate  them  by  a  vertical  4fo26 


line,  and  draw  a  horizontal  line  under  618 

the  dividend. 

And,  commencing  on  the  left,  say :  7  zn  4  is  nought. 

Then  take  together  the  first  two  figures  as  one  number, 
and  say :  7  in  iS,  6  times,  and  1  over  (set  down  the  6 
under  the  hundreds,  of  which  order  it  is) ; 

7  in  12,  1  ti?ne,  and  5  over  (set  down  1  under  the  tens). 

7  in  56,  8  tiines  (set  down  8  under  the  units). 

The  answer  is  as  above,  618. 

REMARKS. 

3.  We  see  in  this  operation  that  we  get  out  suc- 
cessively the  units  of  the  different  orders,  beginning  with 
the  largest,  and  that  the  division  of  thousands,  hundreds^ 
tens,  and  simple  units  by  units,  gives  respectively  in  the 
quotient,  thousands,  hundreds,  tens,  and  units,  or,  in 
general,  that :  The  division  of  a  number  of  units  of  any 
order  by  any  number  of  simple  units,  gives  units  of  the 
same  order  in  the  quotient. 

*  The  divisor  is  more  generally  placed  on  the  left,  and  the  quo- 
tient on  the  right  of  the  dividend.  But  the  position  of  the  divisor 
on  the  right  and  the  quotient  under  it,  is  preferable,  and  I  recom- 
mend it;  because  each  partial  product  is  more  conveniently  made 
by  multiplying  the  divisor  by  a  figure  below  it,  than  across  the 
dividend.  It  has  to  be  so  arranged  for  the  proof  of  division ;  and 
besides,  this  arrangement  is  used  in  algebra  by  ail  modern  mathe- 
maticians. 


52  LESSON   XV. 

4.  In  the  preceding  operations,  we  began  at  the  right, 
because  we  might  expect  to  have  a  large  amount  to  carry 
to  the  next  superior  order.  In  division,  on  the  contrary, 
we  begin  at  the  left  because  we  may  expect  a  remainder, 
which  must  be  joined  to  the  inferior  order  of  units  next 
to  the  right. 

5.  There  is  a  case  which  must  be  particularly  noticed. 
It  is  when  some  orders  of  units  do  not  exist  in  the  quo- 
tient. In  that  case,  the  place  of  the  absent  orders  is 
■occupied  by  zeros. 

OPERATION. 

Take,  for  example,  the  division  of  3256048     by  8 

Since  the  first  figure  3  of  the  order  ZrFo(T« 

of  millions  is  smaller  than  the  di- 
visor 8,  there  can  be  no  millions  in 
the  quotient.     Passing  on  then  to 
the  order  of  hundred  thousands,  we  take  the  first  two 
figures  together,  and  say :  In  32,  how  many  times  8^4 
iirnes^  and  we  set  down  the  4  under  the  hundred  thou- 
sands. 

Then  we  say  :  8  i^i  5  is  0,  and  5  over.  So  that  there 
is  no  ten  thousands  in  the  quotient ;  we  set  0  in  their 
place,  and  say : 

8  in  56,  7  times.    We  set  down  7  under  the  thousands. 

Then^  8  i/i  0,  0 ;  which  we  set  down  in  the  place  of 
hundreds,  to  show  that  there  is  no  figure  of  this  order. 

Again :  4  tens  cannot  be  divided  as  such  by  8.  Wfe 
set  down  0  under  the  tens,  and  divide  48  units  by  8, 
which  gives  6  for  the  last  figure. 

So  that  the  whole  answer  is  407,006,  in  which  the 
figures  4  and  7  get  their  proper  places  and  value  by 
means  of  the  0. 

6.  Sometimes  the  division  cannot  be  exactly  made. 
Let  us  suppose  that  35,681  dollars 
are  to  be  divided  between  6  per- 
sons. 

In  effecting  the  division,  we  find 
a  quotient,  5946,  and  a  remainder, 
5.     Now,  it  is  clear  that  this  re- 


OPERATION. 

35681  I  6 
quotient,  5946 +  f 


LONG   DIVISION.  OS 

mainder,  5  dollars,  is  also  to  be  divided  among  the  6  per- 
sons. This  could  be  done  by  converting  it  into  smaller 
units,  like  cents,  for  instance.  But,  when  this  is  not  re- 
quired, the  remainder  is  written  after  the  quotient^  and 
the  divisor  'placed  under  it,  with  the  sign  of  division 
betioeen  them,  to  shoiv  that  the  remainder  also  is  to  be 
divided. 

Questions. — "What  is  short  division  ?  How  should  the  num- 
ber be  decomposed  ?  How  are  the  numbers  arranged  in  short 
division  ?  Where  do  you  commence  to  divide  ?  Why  ?  Explain 
the  process.  What  is  the  order  of  the  quotient,  when  you 
divide  units  of  any  order  by  simple  units  ?  When  you  divide 
millions?  ten  thousands?  thousands?  hundreds?  &c.,  by  units? 
W^hen  the  units  of  any  order  do  not  contain  the  divisor,  what  is 
done  with  the  quotient  ?  Why  ?  What  is  done  with  the  re- 
mainder ?     Why  ? 

EXERCISES. 

1.  Divide  16,992,360  by  2,  3,  4,  5,  6,  7,  8,  9,  10,  11,  12. 

2.  Also,  132,889,680  by  the  same. 

3.  "  36,00.'),472  by  2,  by  3,  by  4,  by  8,  by  9. 

4.  "  111,111,111  by  3,  by  9. 

5.  «  1,430,400  by    7,=  -        -        -f  Remainder  6i 

6.  «  6,730,214  by  10,=  .         -        4-          «          4. 

7.  «  4,560,389  by    9,  =  -        -     '  +          "          8» 

8.  "  6,080,519  by  12,  = 

9.  «  66,297,439  by  11,= 

LESSON  XVI. 
LONG  DIVISION. 

1 .  Long  Division  is  performed  as  follows : 

I.  Set  down  the  divisor  on  the  right^  of  the  dividend. 
Separate  them  by  a  vertical  line^  and  draw  a  horizontal 
line  under  the  divisor, 

II.  Take  at  the  left  of  the  dividend  just  as  many 
figures  as  will  contain  the  divisor, 

*  See  preceding  note.  The  operation  thus  arranged  is  more 
cdliipact  and  convenient. 

5'^ 


54<  LESSON    XVI. 

III.  Considering  this  number  as  a  jpartial  dividend, 
find  how  often  it  contains  the  divisor^  and  set  the  first 
figure  of  the  quotient  under  the  divisor, 

IV.  Multiply  the  ichole  divisor  by  this  figure,  and 
subtract  the  product  from  the  first  partial  dividend* 

V.  Next  to  the  remainder,  bring  down  the  following 
figure  of  the  dividend;  then  find  how  often  the  divisor 
is  contained  in  this  second  partial  dividend,  and  write 
the  new  quotient  on  the  right  of  the  preceding  one, 

VI.  Multiply  the  tvhole  divisor  by  the  new  quotient, 
subtract  from  the  second  partial  dividend.,  and  bring 
doivn  the  following  figure  of  the  dividend ;  you  have 
then  a  third  partial  dividend, 

VII.  Continue  the  operation  in  the  same  way  until  all 
the  figures  of  the  dividend  have  been  brought  down, 

Vni.  If  there  is  a  remainder,  write  it  after  the  quo* 
tierU,  and  the  divisor  under  it  (XV.,  6).* 

2.  To  explain  the  above  rule,  let  it  be  proposed : 

To  divide  equally  557,793  dollars  between  849  per* 
sons. 

This  amount  may  be  considered,  as  was  done  in  short 
division,  to  be  composed  of  as  many  parcels  as  there  are 
orders  of  units  in  it ;  and  then  we  have  to  divide  each 
parcel  successively. 

The  first  inquiry  is,  what  will 
be  the  largest  order  of  units  or 
parcel  in  the  quotient,  and  how 
many  figures  it  will  contain  ?  It 
is  evident  that  the  quotient  can 
contain  no  hundred  thousands, 
since  there  are  but  5  of  them  for 
849 ;  no  ten  thousands,  because 
there  are  only  55  of  them;  andrio 
thousands,  since  their  number, 
557,  though  composed  of  three 
figures,  does  not  contain  849. 

*  The  division  of  the  remainder  by  the  divisor  thus  indicated,  "la 
generally  written  in  smaller  figures  than  the  quotient. 


OPERATION. 

849 


557,793 
5094 

4839 
4245 

5943 
5943 

0000 


657 


LONG   DIVISION.  55 

Hence,  the  first  part  of  the  number,  which  contains 
*he  divisor,  must  be  composed  of  four  figures,  one  more 
than  the  divisor.  This  first  partial  dividend y  5577, 
being  of  the  order  of  hundreds  (YI.,  7),  it  follows,  that 
the  first  figure  in  the  quotient  must  also  be  of  the  order 
of  hundreds  (XY.,  1  and  8),  and  that  the  quotient  will 
contahi  three  figures  or  orders  of  units,  hundreds^  tens, 
and  units. 

This  first  investigation  is  very  important,  and  should 
always  precede  the  operation,  to  guard  against  mistakes. 

Now,  dividing  the  first  partial  dividend,  5577  hun- 
dreds, by  849,  we  find  for  the  first  share  6  hundreds, 
which  we  set  in  the  quotient. 

Then  the  multiplication  of  6  hundreds  by  the  divisor, 
849,  gives  a  product  of  5094  hundreds,  for  the  first  total 
fihare  or  parcel  to  be  taken  from  5577. 

The  subtraction  leaves  a  remainder,  483  hundreds. 

3.  Having  now  disposed  of  the  hundreds,  let  us  plro- 
ceed  to  ascertain  how  many  tens  the  quotient  will  contain. 
For  this  purpose,  we  must  convert  the  483  hundreds  into 
tens.  We  know  that  they  are  equal  to  4830  tens,  to 
which  the  9  of  the  general  dividend  must  be  added; 
making,  in  all,  4839  tens. 

This  is  the  reason  why  9  is  brought  down  next  to  483. 
Its  annexation,  according  to  the  principles  of  numeration, 
makes  at  once  a  number,  4839,  of  tens  (YL,  7). 

The  second  partial  dividend,  4839  tens,  divided  by  the 
units  of  the  divisor,  will  give  us  5  tens  in  the  quotient ; 
which,  multiplied  by  the  divisor,  make  4245  tens,  the 
second  share  or  parcel  to  be  subtracted  from  the  dividend ; 
leaving  a  remainder,  594  tens. 

By  the  side  of  this,  now,  we  bring  down  the  3  units 
of  the  general  dividend,  and  we  form  the  number  of  5943 
units. 

This  last  partial  dividend  divided  by  849,  gives  exactly 
7  units  in  the  quotient ;  the  7  units,  repeated  849  times 
and  subtracted,  leaving  no  remainder. 

So  that  the  whole  quotient  is  657. 

Questions. — What  is  long  division  ?  Repeat  the  rules.  How 
are  the  numbers  set  down  ?     How  many  figures  do  you  take  lor 


56  LESSON   XVII. 

the  first  partial  dividend  ?  Of  what  order  will  the  first  figure  of 
the  quotient  be  ?  How  many  figures  will  it  contain  ?  Why  is 
the  next  figure  of  the  dividend  brought  down? 

Exercises. — Use  those  given  in  Multiplication. 

LESSON  XVII. 

1.  This  Lesson  contains  some  practical  methods  to 
facilitate  the  operation  of  Division. 

In  the  first  'place y  we  have  implicitly  admitted  that  the 
quotient  of  each  partial  dividend  could  be  readily  ascer- 
tained. But  this  is  not  always  the  case  when  the  divisor 
is  large ;  then, 

The  figure  in  the  quotient  for  each  partial  dividend 
is  found  by  trials  asfolloivs  : 

If  tJie  second  figure  of  the  divisor  does  not  exceed  5, 
see  how  often  its  first ,  or  first  tico  figures  are  contained  in 
the  first  part  of  the  dividend ;  the  result  will  prohahly  he 
the  required  figure  of  the  quotient. 

If  the  second,  figure  of  the  divisor  is  over  5,  add  one 
unit  to  its  first  figure,  and  try  it  with  this  addition, 

2.  The  reason  of  this  is,  that,  as  it  would  seldom  be 
possible  to  tell  at  once  how  often  the  whole  divisor  will 
go  in  a  dividend,  an  approximation  is  obtained  by  taking 
one  or  two  figures  at  the  beginning  of  each  for  the  whole 
number,  since  the  size  of  a  number  depends  chiefly  on 
its  first  figure.  (VI.,  3) 

In  the  annexed  example  we  try  8  in  55 ;  but 
it  must  be  recollected  that  it  is  in  reality  800 
in  5500 ;  and  we  try  8,  because  849  is  nearer 
to  800  than  to  900. 

But  if  the  divisor  were  over  850,  for  example 
879,  its  value  being  then  nearer  to  900  than  to 
800,  the  approximation  would  be  greater  with 
9  than  with  8. 

In  many  cases,  like  the  present,  there 
can  be  no  doubt  as  to  any  figure  in  the 
quotient.  It  is  evident  that  8  will  only  go  6  times ;  for, 
5  multiplied  by  8  would  leave  a  remainder  out  of  55 
nearly  double  8 ;  and  7  would  give  the  product  56. 


OPERATION. 
849 


557793 
5094 


4839 
4245 

5943 
5943 

0000 


657 


LONG  DIVISION.  57 

In  the  second  partial  dividend,  4839,  though  48  is  6 
times  8,  6  is  too  large,  because  the  product  of  6  by  4 
would  give  something  to  be  carried  in  addition  to  48. 

In  the  third  partial  dividend,  5943,  the  trial  with  8  in 
59  gives  evidently  7,  since  the  product  of  6  by  8  taken 
from  59  would  leave  a  remainder  much  larger  than  8, 
while  8  would  give  64, 

3.  But  the  conclusions  are  not  always  so  evident ;  and 
it  frequently  happens  that  the  first  trial  is  not  successful. 
This  is  generally  discovered  after  the  divisor  has  been 
multiplied;  and  it  is  clear  that: 

I.  When  the  product  of  the  divisor  hy  the  quotient  is 
larger  than  the  i^artial  dividend,  the  quotient  is  too  large. 

II.  Wien  this  product,  subtracted  from  the  partial 
dividend  J  docs  not  leave  a  remainder  smaller  than  the  di- 
visor, the  quotient  is  too  small. 

If,  we  had,  for  example,  to  divide  2046  by  256,  we 
might  be  tempted  to  try  9,  and  then  8;  both  of  which 
would  be  found  to  give  products  larger  than  the  divisor. 

If,  applying  the  2d  rule,  we  divide  20  by  3,  instead  of  2, 
then  the  quotient  6  would  give  the  product  1536  and  re- 
mainder 510;  larger  than  the  divisor,  showing  that  6  is  too 
small. 

It  is  true,  that,  since  the  larger  the  parts  of  the  divider]  d 
and  divisor  used  in  the  trial,  the  more  certain  the  quotient 
will  be,  we  might  have  tried  25,  which  in  206  goes  8  times, 
and  thus  escaped  the  trial  of  9;  but,  even  then,  the  apparent 
quotient  8  would  have  failed. 

4.  In  order  to  avoid  these  useless  mutiplications  and 
subtractions,  it  is  always  best,  before  multiplying,  to  verify, 
hj/  short  divisio7iy  the  quotient  figure,  as  follows  : 

Divide  the  dividend  by  the  largest  prohahle  quotient 
figure,  comijaring  successively  each  figure  of  the  result  loith 
the  corresponding  one  of  the  divisor,  until  you  get  a  figure 
either  smaller  or  larger  than  in  the  divisor.  In  the  first 
case,  the  quotient  is  too  large;  in  the  other  it  is  right. 

This  method  has  over  the  common  method  the  advantage 
that  it  is  quicker,  more  certain,  and  does  not  require  to  set 


58 


LESSON   XYII. 


down  any  number,  since  every  figure  you  get  is  the  same 
as  in  the  divisor,  until  you  stop. 

For  instance,  in  the  division  of  1269231  by  168654,  we 
try  9  first,  and  say:  9  in  12,  1  (as  the  divisor). 

9  in  36;  4  (smaller  than  the  second  figure  in  the  di- 
visor). We  are  at  once  certain  that  9  is  too  large,  since 
it  is  contained  in  the  dividend  a  number  of  times  less  than 
the  divisor. 

Trying,  now,  8,  the  division  gives  us  successively  the 
following  figures  of  the  divisor:  1,  5,  8,  6,  5;  but,  for 
the  last,  we  get  3  instead  of  4.  Then,  again,  the  number 
which  should  multiply  8  to  make  the  dividend,  is  smaller 
than  the  divisor ;  consequently  8  is  also  too  large. 

Passing  on  to  7,  and  dividing,  we  get  1,  as  in  the  di- 
visor for  the  first  figure  j  but,  for  the  second,  we  get  8, 
which  is  larger  than  the  second  figure,  5,  of  the  divisor, 
and  shows  at  once  that  7  is  right;  since  a  greater  number 
than  the  divisor  being  contained  7  times,  the  divisor 
itself  must  be  contained. 

5.  Secondly,  A  case  frequently  occurs  in  division, 
which,  though  comprised  in  the  general  rule,  should  be 
noticed  separately.  It  is  when  some  orders  of  units  are 
absent  in  the  quotient,  as  in  this  example : 


2^1089975 
260916 

43486 

6004^^3^^. 

. . .  173975 
173944 

....  31 

In  the  first  place,  it  requires  six  figures  of  the  dividend 
to  contain  the  divisor,  the  last  of  them  being  of  the  order 
of  thousands;  it  follows  that  the  first  figure  of  the 
quotient  will  be  of  the  order  of  thousands^  and  the  quo- 
tient must  have  four  figures. 

Now  in  the  first  partial  dividend,  261089,  because  26 
contains 4  at  the  most  6  times,  and  must  contain  it  more  than 


LONG   DIVISION.  59 

5  times,  since  the  product  20,  of  5  by  4,  would  leave  a 
remainder  much  larger  than  the  divisor,  it  follows, 
without  trial,  that  6  is  the  amount  of  thousands  in  the 
quotient. 

Multiplying,  then,  the  divisor  by  6,  we  have  to  sub- 
tract the  first  share,  260916,  and  find  a  remainder  of  173 
thousands  still  to  be  divided. 

Now  we  have  to  find  the  hundreds  of  the  quotient. 
For  this  purpose,  we  change  the  173  thousands  of  the 
dividend  into  hundreds^  by  annexing  the  figure  9  of  the 
hundreds  (XVL,  3). 

The  second  partial  dividend,  1739  hundreds^  does  not 
contain  the  divisor,  43486;  consequently  we  have  no 
hundreds  in  the  quotient,  and  show  it  by  writing  a  0  in 
the  place  of  hundreds. 

In  order,  now,  to  get  the  tens^  a  third  partial  dividend, 
17397  tens^  is  formed  by  bringing  down  7  tens  next  to 
the  last  number,  and  we  see  at  once  that  this  number  is 
likewise  smaller  than  the  divisor.  Hence,  also,  there  are 
no  tens  in  the  quotient,  which  we  show  by  another  0  in 
their  place. 

Finally,  the  last  partial  dividend  of  units  is  formed  by 
bringing  down  the  5  units :  4  units  are  obtained  in  the 
quotient,  and  a  remainder,  31,  is  left,  whose  division  by 
the  divisor  is  indicated,  as  appears  in  the  result.* 

FinST   REMARK. 

6.  When  the  divisor  is  1,  followed  by  ciphers^  as 
10,  100,  1000,  c^'c,  cut  of  from  the  right-hand  of  the 
dividend  as  many  figures  as  there  are  ciphers  in  the 
divisor.  The  figures  on  the  left  will  he  the  quotient; 
those  on  the  right,  the  remainder, 

JSarampZe.— Divide  9865  by  100.  Ans.  98-^^^. 

The  reason  of  this  is  evident ;  for  each  figure  is  thus 
removed  two  places  to  the  right,  and  becomes  one  hun- 

^  *  That  4  is  the  right  quotient,  is  also  obvious  at  the  mere  inspec- 
tion of  the  number,  since  5  times  4  would  be  larger  than  17,  and 
3  limes  4  would  leave  a  remainder,  5,  larger  than  the  divisor. 


60  LESSON   XVII. 

dred  times  smaller;  that  is,  the  8  hundreds  are  now 
8  units,  and  the  9  thousands  are  9  tens. 

Our  notation  expresses  that  the  remainder,  65,  also 
is  to  be  divided  by  100. 

Nothing  is  more  calculated  to  produce  an  impression 
that  a  person  is  ignorant  of  the  very  elements  of  arith- 
metic, than  to  set  a  regular  operation  in  this  case* 

Division  by  1,  is  evidently  the  number  itself. 

SECOND    REMARK. 

7.  When  the  divisor  is  any  other  number  followed  by 
ciphers,  a  similar  rule  is  given  in  all  arithmetics  ]  but  I 
have  seen  so  many  errors  committed  in  applying  it,  that 
it  had  better  be  deferred  until  we  come  to  Decimal 
Fractions  (XXVII). 

Qiiestion-s.—How  do  you  find  the  figure  in  the  quotient  ?  What 
do  you  do  when  the  second  figure  of  the  divisor  is  5  or  more  ? 
Why  ?  When  the  product  of  the  divisor  by  the  figure  in  the 
quotient  exceeds  the  dividend  ?  And  when  the  remainder  is 
equal  to  or  larger  than  the  divisor,  what  do  you  conclude  ?  What 
is  the  best  mode  of  trial  in  doubtful  cases?  What  do  you  con- 
elude  in  trials,  when  you  come  to  a  figure  smaller  than  that  in 
the  divisor  ?  When  to  one  larger  ?  Why  ?  When  a  partial 
dividend  is  smaller  than  the  divisor,  what  do  you  do  ?  What  is 
done  with  the  remainder  ?  When  the  divisor  is  1,  followed  by 
ciphers,  how  do  you  divide  ?    Why  ? 

EXERCISES. 

1.  13312  eggs  are  to  be  packed  in  52  kegs.  How  many  eggs 
in  each  keg?  A?i.s.  25Q  egg-s, 

2.  A  person  buys  land  for  17575  dollars,  at  37  dollars  an  acre. 
How  many  acres  does  he  buy  ?  Ans,  475  acres, 

3.  What  number,  multiplied  by  422,  will  produce  253622  ? 

A71S.  601. 

4.  A  railroad  receives  666490  dollars  per  year.  How  much 
is  it  a  day  ?  Ans,  1826  dollars, 

5.  The  light  of  the  sun  passes  from  the  sun  to  the  earth,  a 
distance  of  about  95,000,160  miles,  in  8  minutes.  How  much  is 
that  per  second  ?  Ans.  191911  miles, 

6.  The  whole  pay  of  a  number  of  soldiers  is  238080  dollars ; 
each  soldier  gets  48  dollars.     How  many  soldiers  are  there  ? 

Ans,  4960  soldiers. 


NATURE   OF   THE   QUOTIENT.  6f 

LESSON  XVIII. 
NATURE  OF  THE  QUOTIENT. 

1.  This  is  the  place  to  introduce  a  remark  of  some  im- 
portance, in  regard  to  the  nature  of  the  quotient. 

It  has  been  said  in  multiplication  that  the  product  is 
always  of  the  same  nature  as  the  multiplicand  (XL,  7). 

2.  In  division,  the  dividend  is  a  product,  of  which 
the  divisor  and  quotient  are  the  factors.  According  to 
the  nature  of  the  question,  either  of  them  may  be  the 
multiplicand,  and  therefore  of  the  same  nature  with  the 
dividend.     It  follows  that, 

I.  The  quotient  is  of  the  nature  of  the  dividend  when 
the  divisor  is  not. 

II.  But  when  the  dividend  and  divisor  are  of  the  same 
nature,  the  units  of  the  quotient  may  be  anything  else, 

EXAMPLE   OF    THE   FIRST    CASE. 

For  24  dollars  4  yards  were  bought  j  how  much  is 
it  a  yard  ?  Ans.  6  dollars. 

Here  the  dividend  and  divisor  are  different,  and  the 
quotient  necessarily  of  the  nature  of  the  dividend. 

EXAMPLE    OF    THE    SECOND    CASE. 

At  4  dollars  a  yard,  how  many  yards  can  be  bought 
for  24  dollars  ?  Ans,  6  yards. 

Here  the  dividend  and  divisor  are  both  dollars ;  the 
quotient  is  different  from  both. 

3.  It  is  an  error  to  imagine  that  in  the  first  example 
we  divide  dollars  by  yards,  and  in  the  second  dollars  by 
dollars. 

In  both  cases,  after  having  ascertained,  from  the  question, 
of  what  nature  the  units  of  the  quotient  are  to  be,  we  find 
their  number  by  tbe  division  of  abstract  numbers ,  without 
further  reference  to  their  nature. 
6 


6^  LESSON   XVIIl. 

It  is  true  that,  in  the  first  case,  it  would  not  be  pre- 
cisely an  error  to  say  that  we  divide  24  dollars  by  4, 
and  get  6  dollars;  but  it  is  by  4,  and  not  4  yards^  we  di- 
vide.     The  divisor  is  invariably  an  abstract  number. 

In  the  second  case,  however,  it  would  be  a  real  ab- 
surdity to  say  that  we  divide  dollars  by  dollars,  and  get 
yards.  We  want  abstractly  to  find  a  number  of  units, 
designated  by  the  question,  equal  to  the  number  of  times 
that  4  is  contained  in  24,  without  reference  to  their 
nature. 

Singular  as  it  may  appear,  the  question  of  dividing  a 
number  of  units  by  units  of  the  same  kind,  such  as  dol- 
lars by  dollars,  is  by  almost  everybody  answered  by  a 
number  of  units  of  the  same  nature  j  when,  in  fact,  it 
is  rarely  so. 

4.  In  the  few  applications  which  follow  the  17th  les- 
son, this  principle  is  further  elucidated.  In  the  second 
and  sixth,  the  dividend  and  divisor  are  of  the  same 
nature  ;  the  quotient  differs  from  both. 

In  the  others,  the  dividend  and  divisor  being  of  dif- 
ferent kinds,  the  quotient  is  of  the  nature  of  the 
dividend. 

The  third  question  relates  only  to  abstract  numbers. 

It  may  be  further  remarked  that,  in  practice,  no  ques- 
tion in  multiplication  or  division  is  complete  unless  three 
quantities  are  given,  one  of  which  is  a  single  unit,  whose 
nature  serves  to  determine  that  of  the  answer.  In  each 
of  the  preceding  applications,  this  unit  will  readily  be 
discovered.* 

PROOF    OF    DIVISION. 

5.  Multiply  the  quotient  hy  the  divisor,  and  to  this 
product  add  the  remainder,  if  there  be  any.  The  sum 
should  be  equal  to  the  dividend, 

*  Multiplication  and  division  are  in  fact  rules  of  three,  in  which 
one  of  the  terms  of  the  proportion  is  a  unit  of  the  nature  of  one  of 
the  numbers. 


PROOF  OF  DIVISION  AND   MULTIPLICATION. 


63 


EXAMPLE. 

PROOF. 

Dividend,     5,778 

25  divisor 

231     quotient. 

60 

.77 

2312% 

quotient. 

25     divisor. 
1155 

75 

462 

.28 

-f-  3     remainder. 

55 

5,778  =  dividend. 

Kemainder        .3 

It  is  evident  that,  since  the  divisor  is  taken  from  the 
dividend  a  number  of  times  equal  to  the  quotient  231, 
and  an  excess  3  is  left,  the  repetition  of  the  divisor  231 
times,  adding  in  the  remainder,  must  reproduce  the  divi- 
dend. 

And,  because  it  makes  no  diflference  which  of  two  factors 
is  taken  for  the  multiplier,  we  prefer  the  divisor,  in  order 
that,  in  making  new  products,  we  may  more  certainly  detect 
errors.,  which  going  over  the  same  multiplication  would 
not  be  as  likely  to  do. 

■-  6.  It  must  be  observed,  that  ^V  is  intrinsically  a  part 
of  the  quotient,  and  that  we  virtually  multiply  by  that 
also  when  we  add  3 ;  for  3  divided  by  25,  is  the  same  as 
the  25th  part  of  3,  and  a  25th  part  repeated  25  times, 
must  make  the  whole  5  that  is,  3. 

PROOF    OF    MULTIPLICATION. 

7.  Division  is  the  most  certain  proof  of  multiplication. 
Divide  the  product  by  either  the  multiplicand  or  multi- 
plier  ;  the  other  must  be  the  quotient.  For,  either  factor 
expresses  how  often  the  other  is  contained  in  the  product. 

EXAMPLE, 

Multiplicand,     2564 


Multiplier, 


Product, 


47 


17948 
10256 


120508 
94 
265 
235 
300 
282 
188 
183 

000 


47     divisor  =  multiplier. 
2564    quotient  =  multiplicand. 


64  LESSON   XIX. 

Questions. — Of  what  nature  is  the  quotient,  when  the  dividend 
^nd  divisor  are  of  different  natures  ?  When  they  are  alike  ? 
Give  examples.  How,  in  the  operation,  are  those  numbers,  and 
especially  the  divisor,  considered?  How  many  quantities  must 
be  known  in  multiplication  and  division,  to  determine  the  nature 
of  the  answer  ?  How  do  you  prove  division  ?  Why  ?  How 
do  you  prove  multiplication  by  division  ?  Why  ?  Is  there  any 
analogy  between  these  operations  and  the  rule  of  three  ?  (See 
note.) 

LESSON  XIX. 

QUICK  DIVISION. 

1.  After  the  student  has  acquired  sufficient  facility  in 
division,  he  may  greatly  abbreviate  it,  both  as  regards 
time  and  space,  by  subtracting  at  the  same  time  that  he 
multiplies. 

RULE. 

Multiply  successively,  by  the  figure  in  the  quotient, 
each  figure  of  the  divisor,  and  subtract  the  product  from 
the  corresponding  figure  in  the  dividend. 

If  the  product  is  larger  than  the  figure  in  the  divi* 
dend,  subtract  this  product  from  the  next  number  above 
it,  ending  with  the  figure  in  the  dividend;  and  then 
carry  the  tens  thus  borrowed  to  the  product  of  the  quO' 
tient  by  the  next  figure  of  the  divisor,  to  be  subtracted 
together  with  this  product,  and  go  on  with  the  operation 
in  the  same  way  to  the  end. 


EXAMPLE. 


557793 
4889 
5943 


849 

657 


Say,  in  55  how  many  times  8  5 
6  times,  and  set  6  in  the  quotient. 

6  times  9  are  54;  from  57 
(which  is  the  next  number  above 
54,  ending  with  the  figure  7)  leaves 
3,  and  5  to  carry.     Set  down  3. 

6  times  4  are  24,  and  5  carried,  are  29 ;  2^  from  37, 
leave  8,  and  3  to  carry.     Set  down  8. 

6  times  8  are  48,  and  3  carried,  51 3  from  55,  leave  4. 
Set  down  4. 


QUICK    DIVISION.  65 

Now  bring  down  9,  and  say  :  8  in  48 ;  it  goes  only  5 
times.     Set  down  5  in  the  quotient,  and  proceed  as  before. 

5  times  9  are  45 ;  from  49  leave  4,  and  4  to  carry , 
5  times  4  are  20,  and  4  carried,  24  j  24^  from  33,  leave  9, 
an<i  3  ^o  carry,  &c.,  to  the  end.  The  final  quotient  is 
657. 

2.  This  method  of  subtracting  and  carrying,  is  only 
an  extension  of  that  used  in  subtraction.  We  have  to 
subtract  the  first  product,  54,  from  7j  and,  in  order  to 
render  the  subtraction  possible,  we  take  the  number  end- 
ing with  7,  next  above  54,  which  is  57,  and  subtract. 

But,  in  doing  so,  we  have  virtually  added  50  to  the 
figure  7  of  the  partial  dividend  :  and,  in  order  not  to  alter 
the  remainder,  we  must  also  add  the  same  amount  to  the 
number  we  subtract,  which  we  do,  as  5  units  of  the  next 
higher  order  joined  to  the  product  24. 

Thus  we  get  29:  and,  as  this  cannot  be  subtracted  from 
3,  we  take  the  number  ending  in  3  next  above  it,  which 
is  33,  and  subtract,  &c. 

A  little  practice  will  make  this  method  familiar,  and  it 
will  prove  much  more  convenient  to  a  correct  calculator. 

It  will  be  used  exclusively  in  subsequent  examples. 

Questions. — In  what  does  quick  division  differ  from  common 
division  ?  How  do  you  subtract  ?  What  number  do  you  subtract 
from  ?  What  do  you  carry  ?  Give  the  demonstration.  Repeat 
the  rule.     What  is  the  advantage  of  this  method  ? 

EXAMPLES   FOR   PRACTICE   IN   DIVISION. 

1.  8,760    :   24  =  2.^^, 

2.  7,629    :   32  = +  1 1. 

3.  40,200    ;   75  =  536. 

4.  274,036    :   48  =  .         .         .         .        +4  Rem. 

5.  257,355    :   35  =  7,353. 

6.  599,359    :   78  =  .         .         .         .        +7  Rem. 

7.  4,280,822    :    91  =  47,042. 

8.  750,969   :   75  =  .         .         .         .         +69  Rem. 

9.  5,616,072    :   84  =  66,858. 

10.  1,195,030   :   99  =  .         .         .         .         +1  Rem. 

6  *  E 


66 


LESSON  XIX. 


11.  15,076,944  : 

12.  12,864,175  : 

13.  46,508,928  : 

14.  999,999,999 


72  =  209,402. 

32  = 

96  =  484,468. 

;  81  = 


15.  252,144  :  412  =  612. 

16.  136,714  :  534  = 

17.  2,777,848  :  487  =  5,704. 

18.  807,000  :  394  = 

19.  511,425,396  :  5,847  =  87,468. 

20.  405,768,567  :  50,406  = 

21.  436,940,074  :  64,237  =  6,802. 

22.  9,638,789  :  3,456  = 

23.  270,974,414,340  :  276,390  =  980,406 

24.  33,188,281,929  :  31,456  = 

25.  103,805,467  :  38,446  = 

26.  57,824  :  10  = 

27.  694,359  :  100  = 

28.  89,045,000  :  1000  =' 

29.  6,472,001  :  1000  = 

30.  573,070,698  :  6,587  = 

31.  921,253,442,978,025  :  918,273,645  = 

32.  435,603,662,775    :   870,496  = 

33.  352,107,028,000    :    1,672,940  =  210,472 


15  Rem. 


10  Rem. 


88  Rem. 


+    267  Rem. 


+        5  Rem. 


+ 


9  Rem. 


4-  1267  Rem. 


4-  1698  Rem. 
+  500,407  Rem. 

_, 32  0 

"t"    1^7  2y40* 


The  examples  with  answers,  may  also  serve  for  multiplica- 
tion. 

Those  without  answers  are  intended  for  exercise  on  some  par- 
ticular difficulty,  which  would  be  removed  to  the  disadvantage 
of  the  student,  if  the  quotients  were  given. 

The  above  examples  may  be  varied  by  changing  remainders 
In  the  beginning,  those  given  in  multiplication  may  be  used. 

PROOF   OF   QUICK   DIVISION. 

In  proving  quich  division,  you  may  multiply  the  divisor 
by  tlie  quotient;  as,  in  this  case,  there  is  not  the  objection 
noticed  in  Lesson  XVIII.,  5;  since  here  the  partial  pro- 
ducts are  not  written  down. 


DECIMAL   FRACTIONS.  87 

CHAPTER  in. 

CONTAINING   DECIMAL    FRACTIONS. 

LESSON  XX. 
DECIMAL  FRACTIONS. 

SUPPLEMENT    TO    NUMERATION. 

1.  In  the  preceding  lessons  we  have  considered  only 
whole  numbers  or  integers  ;  that  is,  numbers  composed  of 
whole  units. 

This  lesson  will  introduce  the  subject  of  fractions. 

2.  The  word  Fraction  implies,  a  part  of  any  thing, 
the  meaning  is  only  relative  to  that  thing  ;  for,  a  fraction 
of  a  thing  may  be  a  whole  number,  in  regard  to  another. 
For  example : 

A  te7i  cent  piece  is  a  fraction  of  a  dollar ;  it  is  a 
whole  number,  when  compared  to  one  cent, 

A  company  of  men  is  a  whole  number  of  meii ;  it  is  a 
fraction  of  a  battalion. 

The  battalion  itself  is  a  fraction  of  the  regiment,  and 
that  only  a  fraction  of  the  army. 

An  inch,  a  foot,  a  yard,  are  all  units;  yet  the  inch  is 
z,  fraction  of  the  foot^  the  foot  of  a  yard^  and  the  yard 
itself  of  a  mile.     They  are  all  relative  units. 

3.  Decimal  fractions  (or  simply  decimals)  are  those 
which  express  a  division  of  any  unit  into  ten  equal  parts. 

4.  They  are  formed  by  an  extension  of  the  decihial 
system  of  numeration  and  notation  to  a  descending  scale 
below  simple  units. 

We  have  seen,  in  numeration,  that  units  increase  in 
value  ten  times  for  every  place  they  are  removed  to  the 
left. 

Evidently,  therefore,  every  order  of  units  is  ten  times 
smaller  for  every  place  that  it  stands  removed  from  the 
one  on  its  left. 


68 


LESSON   XX. 


Thus,  while  one  thousand  is  ten  times  larger  than  one 
hundred,  this  is  ten  times  smaller  than  one  thousand. 

Again,  one  thousand  is  equal  to  one  hundred  tens^ 
hence,  one  ten,  the  tenth  part  of  a  hundred,  is  the  hun- 
dredth part  of  a  thousand,  &c. 

The  unit  is  the  tenth  part  of  one  ten,  the  hundredth 
part  of  one  hundred,  the  thousandth  part  of  one  thousand. 

Viewed,  therefore,  in  reference  to  superior  units,  infe- 
rior ones  are  truly  decimal  parts. 

But  there  is  nothing  to  prevent  us  from  carrying  the 
descending  scale  by  tenths^  below  the  simple  unit  of  com- 
parison^ and,  upon  the  same  principle,  to  consider 
The   1st   order   to   the   right   of  simple  units   as   com- 
posed -         of  tenths  of  that  unit. 
The  2d 5             -         -         of  hundredths* 
The  3d,             -         -         of  thousandths. 
The  4th5            -         -         of  tenths*  of  thousandths. 
The  5th,            -         -         of  hundredths  of  thousandths^ 

&c.  &c. 

5.  The  corresponding  notation  exhibited  in  the  follow- 
ing table,  is  as  simple  as  the  principle  itself: 


Ten  millions,  &c. 

20000000. 

Millions 

2000000. 

Hundred  thousands 

200000. 

Ten  thousands 

20000. 

Thousands 

2000. 

Hundreds 

200. 

Tens 

20. 

Units 

S. 

.2 

tenths. 

.02 

hundredths. 

.002 

thousandths. 

.0002 

tenths  of  thousandths. 

.00002 

hundredths  of  thousandths. 

.000002 

millionths. 

.0000002 

tenths  of  millionths,  &c. 

*  There  exists  frequently  some  ambiguity  as  regards  the  reading 
of  decimal  fractions.  In  some  approved  Arithmetics,  we  find  ten 
thousandths  ;  in  others,  tens  of  thousandths.  Both  of  which  I  think. 
incorrect ;  for  the  unit  is  certainly  one-tenth  of  one-thousandth^  but 


NUMERATION    OF    DECIMALS.  69 

6.  The  ascending  and  descending  scales  are  separated 
by  a  dot,  usually  called  the  decimal  point,  but  which  I 
believe  it  would  be  better  to  call  the  U7iits'  point,  I  shall 
frequently  use  this  expression.*  It  is  placed  to  the  right 
of  the  units  and  before  the  tenths, 

7,  The  value  of  decimal  units,  like  that  of  whole  numr 
hers,  depends  on  the  place  they  occupy  ;  that  is,  on  their 
distance  from  the  simple  units. 

In  order  to  fix  the  place  of  deci?nal  figures,  Os  are 
made  to  occupy  the  place  of  absent  orders. 

The  successively  decreasing  decimals  in  the  diagram 
explain  this  extension  of  the  system  of  notation. 

We  can  now  read  with  ease  any  simple  decimal  frac- 
tion :     0.4        or  .4        is  4  tenths. 

0.06      or  .06      is  6  hundredths. 
0.009    or  .009    is  9  thousandths. 
0.0005  or  .0005  is  5  tenths  of  thousandths,  &c. 
The  0  in  the  place  of  units  is  at  pleasure  used  or  omit- 
ted, since  the  decimal  or  units'  point,  made  very  distinct, 
is  sufficient  to  fix  the  place  of  each  order  of  figures. 

Questions. — What  is  a  fraction  ?  Is  a  fraction  absolute  or 
relative  ?  Give  examples.  What  are  decimals  ?  How  do  the 
figures  of  decimals  compare  with  each  other  ?  May  a  whole 
number  be  considered  as  a  fraction  of  another  ?  Give  examples. 
What  is  the  descending  decimal  scale  ?  The  1st,  2d,  3d,  &c., 
order  ?  How  are  whole  numbers  separated  from  decimals  ?  How 
is  any  order  indicated  when  there  are  absent  units?  Read,  0.05; 
0.00002 ;  0.0000003,  &c.  Is  0  indispensable,  when  simple  units 
are  absent  ?    Why  ? 

LESSON  XXL 

1.  Let  us  now  read  a  decimal  number  composed  of 
several  figures ;  for  example, 

not  ten  times  one  thousandth.  Besides,  how  could  400  thousandths  in 
this  way  be  distinguished  from  4  hundred  thousandths  ?  The  read- 
ing, 4  hundredths  of  thousandths,  is  not  only  correct,  but  removes 
all  ambiguity. 

*  On  the  continent  of  Europe,  the  comma  is  used  to  separate 
decimals,  and  a  dot  to  divide  periods.  In  the  United  States,  the 
decimal  point  is  generally  adopted,  and  the  comma  is  employed  to 
divide  the  periods  of  numbers.  I  think  this  notation  preferable  to 
the  European. 


70  LESSON   XXI. 

0.0084593 


This  number  contains : 


+2      ±i      tM        +J      ^       tM 


-£   -^     §   :S    -C    ;Z5 


^  -5         -S  ,^ 


O    O    CO    Ti<     >0    05     CO 


But  the  remark  of  (VI.,  7)  enables  us  to  read  it  off  as  a 
single  number,  by  giving  it  the  denomination  of  the  last 
figure,  without  naming  the  others;  that  is,  eighty-four 
thousand  five  hundred  and  ninety-three  tenths  of  mil- 
lionths  (84<,593  tenths  of  millionths). 

For,  the  successive  figures  stand  in  regard  to  the  respec- 
tive value  of  their  units,  according  to  the  decimal  scale, 
and  therefore  may  be  considered  as  one  number. 

2.  Sometimes  a  whole  number  is  accompanied  by  deci- 
mals. It  is  then  called  a  mixed  or  fractional  number  ; 
as,  for  example,  43.29. 

Which  may  be  read,  either  43  units  and  29  hundredths, 
or  4,329  hundredths ;  for,  it  is  evident  that  the  decimal 
relation  exists  between  all  these  figures, 

9  being  a  number  of  units  of  the  order  of  hundredths. 

2  is  equal  to  2  tens  of  them, 

3  ''        to  300  of  them, 
and  4       '^        to  4000  of  the  same. 

N.  B. — When  an  abstract  decimal  number  is  read  into 
two  parts,  the  word  unit  should  never  be  omitted  after 
the  whole  number;  otherwise  there  would  sometimes  be 
ambiguity. 

For  example,  how  could     200.004     be  distinguished 
from     -  0.204 1 

both  reading  two  hundred  and  four  thousandths;  unless, 
after  200,  the  word  units  be  introduced,  and  we  read  200 
units  and  4  thousandths^  which  removes  all  doubts. 


NUMERATION   OF    DECIMALS.  71 

3.  In  the  case  of  denominate  numbers,  the  denomina- 
tion being  always  given  after  the  units,  there  can  never 
be  any  uncertainty.  If  they  were  yards,  for  instance, 
we  would  read 

200  yards  and  4  thousandths  (of  a  yard). 

4*.  The  writing  of  decimal  fractions  does  not  present 
any  greater  difficulty  than  the  reading : 

I.  Write  the  decimal  number  as  it  is  spoTcen,  just 
as  you  would  write  a  whole  number  /  and  cut  off  by  the 
units'^  or  decimal  point  as  many  figures  as  the  denomina^ 
tion  of  the  last  figure  requires. 

II.  If  there  be  not  as  many  figures,  complete  their 
number  by  ciphers  between  the  first  significant  figure 
and  the  decimal  point, 

EXAMPLES. 

1st.  Write  two  hundred  and  thirty  four  hundredths  of 
thousandths. 

The  denomination  being  of  the  5th  order  of  decimals, 
we  must  have  five  decimal  figures. 

We  write,  first,  234<,  and,  to  move  4  to  the  5th 
place,  we  prefix  two  Os,  and  then  the  decimal  point, 
which  gives  -        0.00234,  or  .00234. 

2d.   Write  9  thousand  and  35  hundredths. 

Set  down,  first,  9035,  and,  because  the  last  denomi- 
nation is  hundredths^  cut  off  two  figures,  showing  that 
the  number  is  90.35 ;  equivalent  to  90  whole  units  and 
35  hundredths. 

5.  The  first  example  shows  that  the  introduction  of 
each  0  between  the  decimal  point  and  the  significant  part 
of  a  fraction,  reduces  it  ten  times  j  for,  it  removes  all  its 
figures  one  place  to  the  right. 

6.  In  general,  in  numbers  attended  with  decimals,  the 
position  of  the  units'  point  determines  the  place  of  the 
entire  units;  and  hence  the  place  and  value  of  every 
significant  figure.     Consequently, 

I.  A  number  is  increased  ten  times  for  each  place  the 
vnits^  point  is  removed  to  the  right. 


72  LESSON   XXI. 

11.  A  number  is  decreased  ten  times  for  each  place  the 
units'^  point  is  removed  to  the  left. 

Since,  in  the  first  case,  each  figure  ascends  as  much, 
and,  in  the  second,  descends  as  many  places. 

7.  Zeros,  in  both  cases,  are  added,  when  the  number 
of  fio;ures  is  not  sufficient. 


23.56 
increased      10  times,  ==        235.6 
100  times,  =      2356. 
1000  times,  =    23560.  with  the  addition  of  0. 
10,000  times,  =  235600.  with  two  Os,  &c. 

23.56  decreased  10  times,  =  2.356 
100  times,  =    .2356 
1000  times,  =    .02356, with  one  0  prefixed. 
10,000  times,  =    .002356,  with  two  Os,  &c. 

These  examples  show  the  great  importance  of  putting 
the  decimal  point  in  its  right  place.  It  is,  indeed,  the 
only  difficulty  in  operations  with  decimals. 

8.  Another  very  important  principle  is  the  following : 
The  addition  of  any  number  ofOs  after  a  decimal fraC' 

lion  does  not  alter  its  value. 

For,  it  is  clear  that,  so  long  as  the  position  of  the  units* 

point  is  not  changed^  the  value  of  all  the  figures  also 

remains  unchanged. 


«       3 


For  example,  to  9.75 


•et  us  add  several  Os ;  thus,      9.7  50000. 


NUMERATION    OF    DECIMALS. 


ti 


The  9  which  precedes  the  units'  point,  remaining  units, 
the  7  remains  tenths^  and  the  5  hundredths,  as  before. 

The  reading  only  of  the  number  is  changed  from  9.75 
hundredths  to  9,750,000  millionths,  which  is  the  same 
thing,  since  9  units  =  9,000,000  millionths. 

And  75  hundredths  =  750  thousandths  —  7500  tenths 
of  thousandths  =  75,000  hundredths  of  thousandths  = 
750,000  millionths. 

In  other  words,  the  addition  of  four  Os  makes  the  units 
of  the  fraction  10,000  times  smaller,  but  compensates  it 
by  increasing  their  number  10,000  times. 

9.  Conversely,  zeros  at  the  right  hand  of  a  decimal 
fraction  may  be  struck  out  without  altering  its  value, 

Q^iestions. — How  do  you  read  decimal  fractions  ?  Why  ?  What 
two  ways  are  there  to  read  a  number  accompanied  by  decimals  ? 
Why  ?  What  should  not  be  omitted  when  read  in  two  parts  ? 
Why?  How  do  you  write  decimals  ?  What  do  you  do  when  the 
figures  of  the  decimal  number  are  not  as  many  as  the  places  of 
decimals  ?  What  is  the  effect  of  each  0  placed  after  the  units' 
point,  before  the  significant  part  of  decimals  ?  What  is  the  effect 
of  moving  the  decimal  point  to  the  left  I  To  the  right  ?  Why  ? 
How  do  you  increase  a  number  10,  100,  1000  times,  &c.?  How 
do  you  decrease  it  10, 100, 1000,  &c.,  times  ?  What  is  done  in  either 
case,  when  the  number  of  its  figures  is  not  equal  to  the  requisite 
places  ?  What  effect  have  zeros  added  to  the  right  of  a  decimal 
fraction  on  its  value  ?  On  its  reading  I  Why  ?  What  effect  will 
the  striking  out  of  zeros,  at  the  end  of  a  decimal  fraction,  have 
on  its  value  ? 


EXERCISES. 


Read  and  write  in  words  : 


.6 

.9661 

.40789 

.910457 

.2100007 

.29 

.0753 

.10101 

.061042 

.0102003 

.04 

.0107 

.60606 

.002004 

.00001072 

.106 

.0025 

.02149 

.030303 

.0000000209 

.251 

.0016 

.00301 

.000561 

.0000600007 

.019 

.0009 

.00036 

.000022 

.0000000008 

.002 

.3472 

.00007 

.000005 

.00000000017 

.090 

.0106 

.01009 

.030034 

.00700009005 

Read  both  ways,  viz.,  as  whole  and  as  mixed  numbers  : 


45.05 

3000.0009 

400.036 

70.007 

.3009 

.436 

200.006 

24.0008 

6007.001001001 

0.206 
7 

3650.009008 

805.0009009 

74f  LESSON   XXI. 

Write 

1.  Six  tenths. 

2.  Four  tenths  and  six  hundredths. 

3.  Nine  hundred  and  nine  thousandths. 

4.  One   thousand,   five    hundred   and    fifty-seven  tenths   of 
thousandths. 

5.  Nine  hundred  and  eleven  hundredths  of  thousandths. 

6.  Fifty-eight  tenths  of  thousandths. 

7.  Four  hundred  and  four  hundredths  of  thousandths. 

8.  Six  thousand  and  seven  millionths. 

9.  Eight  hundred  thousand  two  hundred  and  seven  millionths. 

10.  Six  thousand  two  hundred  and  five  tenths  of  millionths. 

11.  Four  hundred  and  three  millionths. 

12.  Six  hundred  and  six  thousand  and  six  millionths. 

13.  Seven  million  two  hundred  and  fifteen  thousand  and  eight 
tenths  of  millionths. 

14.  Ten  units  and  two  hundredths. 

15.  Two  hundred  and  one  units  and  twenty-five  thousandths. 

16.  Six  hundred  and  forty-four  tenths. 

17.  Nine  thousand  and  ninety-nine  thousandths. 

18.  Seven  hundred  thousand  units  and  seven  hundredths  of 
thousandths. 

19.  Five  million  six  hundred  and  sixty-seven  thousand  and 
eighty-seven  tenths  of  thousandths. 

20.  Two  million  three  hundred  units  and  fifty-four  tenths  of 
millionths. 

21.  Sixty  millions  and  five  millionths. 

22.  Twelve  billions  of  millionths. 

23.  Sixty-four  units  and  seven  hundredths  of  thousandths. 

24.  Six  thousand  units  and  two  hundred  and  three  tenths  of 
thousandths. 

25.  Six  thousand  two  hundred  and  three  tenths  of  thousandths. 

1.  Increase  63.756  ten  times,  one  hundred  times,  one  thousand 
times. 

2.  Increase  .6897,  10 ;  100;   1000;  10,000;  100,000  times. 

3.  Multiply  4.652  by  10;  100;   1000;  10,000;   100,000. 


ADDITION   AND   SUBTRACTION   OF    DECIMALS.  75 

4.  Increase  .05  ten  times,  .06  hundred  timeSj^.O?  one  thousand 
times. 

5.  Multiply  .006  by  10;  .007  by  100;  0.005  by  1000;  .003  by 
10,000 ;  0.004  by  one  million. 

6.  Increase  .000009,  one  hundred  thousand  times. 

«         .00005,  one  million  of  times. 

7.  Decrease  25.72,  10 ;   100;   1000;   10,000  times. 

8.  «  .69,  10;   100;   1000  times. 

9.  Divide  1.05  by  100  ;  2.62  by  one  thousand. 
10.        «     0.007  by  ten ;  0.0008  by  ten  thousand. 

/ 

LESSON  XXII. 
ADDITION  AND  SUBTRACTION  OF  DECIMALS. 

1.  The  principles  that  only  units  of  the  same  kind  can 
be  added  or  subtracted,  applies  evidently  to  descending 
as  well  as  to  ascending  orders.     Therefore, 

In  order  to  add  or  subtract  numbers  with  decimals^ 
set  down  the  numbers  so  that  the  units'  points  may  all  he 
in  the  same  column,  ivhich  will  place  units  of  the  same 
order  under  each  other. 

Then  add  or  subtract,  as  with  whole  numbers;  be 
carefid  to  place  the  decimal  point,  in  the  result,  between 
the  units  and  tenths. 

N.  B. — In  both  operations,  it  will  be  convenient,  though  not 
indispensable,  to  equalize  the  number  of  decimals  ;  that  is,  by  the 
addition  of  zeros  to  make  the  number  of  decimal  figures  of  all 
the  numbers  equal  to  the  greatest.  This  we  are  authorized  to 
do  by  the  remark  of  Lesson  XXL,  8. 

EXAMPLE    IN   ADDITION. 

To  add  29.0146  +  3,146.5  +  2,109  +  .62417  +  14.16: 

Without  equalizing  decimals.  The  decimals  equalized. 

29.0146  29.01460 

3146.5  3146.50000 

2109.  2109.00000 

.62417  .62417 

14.16  14.16000 


5299.29877  5299.29877 


76 


LESSON   XXII. 


In  long  additions,  the  second  arrangement  is  clearly 
less  liable  to  mistakes. 


EXAMPLES   IN   SUBTRACTION. 


Not  equalized. 

1.9185 
-  .625 

=  1.2935 

214.81 
—   4.90142 

=  209.90858 


2,714. 


(!•) 


(2.) 


(3.) 


.00916 


Equalized. 

1.9185 
-  0.6250 

=  1.2935 

214.81000 
—   4.90142 

=  209.90858 

2,714.00000 
0.00916 


=  2,713.99084  =2,713.99084 

The  second  arrangement  is  particularly  recommended 
to  beginners. 

Questions. — How  do  you  add  numbers  with  decimal  fractions? 
How  do  you  subtract  ?  How  do  you  arrange  the  numbers  ?  Why  ? 
Where  do  you  place  the  decimal  point  in  the  result  ?  What  is 
meant  by  equalizing  the  decimals  ?    What  is  the  advantage  of  it  ? 


EXERCISES. 

1.  376.25  +  7.125  +  9762.0047  +  .62  +  77.0005  +  41.  = 
10,264.0002. 

2.  3.5  +  47.25  -|-  2.0073  +  927.01  +  1.5  +  .7327  =  982. 

3.  276  +  54.321  +  112  +  0.65  +  12.5  +  .0463  =  455.5173. 

4.  235  +  .00092+2.0415  +  .6  +  .07  +  .00018  --=  237.7126. 

5.  Add  five  units  and  nine  tenths ;  sixty-nine  hundredths  ;  two 
units  and  eleven  thousandths  ;  twelve  units  and  six  hundredths  ; 
seventy-five  thousand  and  four  tenths.  Ans.  7521.061. 

6.  Add  two  hundred  and  two  units  and  twelve  millionths;  one 
hundred  thousand  and  three  tenths  of  thousandths ;  forty-five 
units  and  ninety-nine  thousandths;  two  hundred  and  sixty-five 
units  and  six  hundredths  of  thousandths  ;  ten  million  six  hundred 
and  twenty-five  tenths  of  thousandths.  Ans,  1;522. 161872. 


SUPPLEMENT   TO   MULTIPLICATION.  77 

7.  62.09  —  23.0784=  8.     10.25  —  2.4512  = 

9.     8  — .006714=  10.       2  — .000001  = 

11.  25  —  0.0091  =  12.     15.1  —  14.99999  = 

13.  From  1.05  take  19  tenths  of  thousandths. 

14.  From  6.0001,  take  23  hundredths  of  thousandths. 

15.  From  2  thousandths,  take  2  millionths. 

16.  From  one  unit,  take  one  hundred  and  eleven  tenths  of 
thousandths. 

17.  From  one  thousand,  take  one  thousandth. 

18.  From  one  hundred  thousand  units,  take  one  hundredth  of 
thousandths. 

19.  From  two  hundred  units  and  five  thousandths,  take  two 
hundred  and  five  thousandths. 

20.  From  two  hundred  tenths,  take  one  hundred  and  ninety- 
nine  thousandths. 

LESSON  XXIII. 

SUPPLEMENT    TO    MULTIPLICATION. 

1.  It  is  necessary  lo  introduce  here  some  additional 
considerations  in  regard  to  multiplication  and  division,  to 
facilitate  the  understanding  of  these  rules,  as  applied  to 
decimal  fractions. 

PROPOSITION   I. 

I.  If  the  multiplicand  or  multiplier  he  made  a  certain 
numher  of  times  larger  or  smaller.^  the  product  will  he 
larger  or  smaller  the  same  numher  of  times. 

In  the  first  place,  as  regards  the  multiplicand,  we  know 
that  the  product  is  equal  to  as  many  multiplicands  as  there 
are  units  in  the  multiplier ;  and,  of  course,  if  there  are 
several  such  multiplicands  to  be  repeated,  the  repetition 
of  each  giving  the  original  product,  the  aggregate  will  be 
equal  to  as  many  products  as  there  are  multiplicands. 

When  I  multiply  4  by  5,  it  is  the  addition  of  4  five  times,  and 
the  product  is  20.  Bat,  now,  if  I  made  the  multiplicand  4 
3  times  larger,  it  would  be 

4-4-4  +  4, 
which,  repeated  5  times,  would  be  4  4-  4  -j-  4 

4  +  4  +  4 
4  +  4  +  4 
4  +  4  +  4 
4  +  4  +  4 

.y4>  =20  +  20  +  20 


78  LESSON   XXIII. 

the  whole  amount  being  equal  to  3  times  the  original  product  ,' 
that  is,  as  many  times  as  the  number  of  multiplicands. 

If  there  are  fewer  multiplicands,  there  is  less  to  repeat, 
and  the  product  must  be  smaller  in  proportion. 

For  example^ — I  give  4  apples  to  each  of  3  boys ;  it  is 
in  all  the  multiplicand  4  repeated  3  times,  r=  12  apples. 

If,  instead  of  this,  1  were  to  give  but  2  to  each,  the 
new  product  would  be  the  same  as  if  I  took  away  one- 
half  of  the  share  of  each  ;  which  would  reduce  the  whole 
amount  to  one-half  of  12,  or  to  6  apples. 

2.  Secondly,  as  regards  the  multiplier,  the  principle 
may  be  understood  to  be  the  same  from  the  remark 
(XIL,  5)  that  if  it  be  taken  for  the  multiplicand,  the  pro- 
duct is  the  same. 

Or,  by  considering  that  the  product,  being  equal  to  as 
many  multiplicands  as  there  are  units  in  the  multiplier,  the 
increase  or  reduction  of  this  number  of  units,  produces  a 
corresponding  increase  or  reduction  in  the  product. 

If,  for  instance,  I  have  to  give  4  apples  to  each  boy  that  pre- 
sents himself,  I  would  give  12  apples  to  3  boys ;  but,  if  5  times 
this  number  of  boys  were  to  come,  I  should  have  to  do  the  same 
thing  5  times ;  that  is,  to  give  5  X  12  =  60  apples. 

3.  Hence,  ivhen  more  convenient,  we  may  multiply  or 
divide  the  product  instead  of  its  factors. 

EXAMPLES. 

I.  I  buy  125  yards  of  cloth,  at  4  dollars  ==  500  dollars.  Now, 
I  want  7  times  that  quantity ;  instead  of  multiplying  7  by  125, 
and  then  by  4,  I  observe  that  4  times  125  is  500,  and  I  multiply 
at  once  500  X  7  =  3,500  dollars. 

II.  Or  else,  I  wish  to  change  it  for  an  equal  quantity,  7  times 
dearer.  Here,  again,  instead  of  multiplying  4  by  7,  and  then  by 
125,  I  say  at  once  7  X  500  =  3,500  dollars. 

PROPOSITION    II. 

4.  If  one  of  the  factors  be  mtdtiplied  by  a  certain 
number,  and  the  other  be  divided  by  the  same,  the  pro- 
duct will  remain  unchanged 


*  SUPPLEMENT    TO    DIVISION.  79 

This  practically  useful  principle  is  almost  evident  from 
what  precedes,  since  one  operation  destroys  the  other. 
Instead  of  buying  6  yards,  at  4  dollars,  you  prefer  to  buy  only 

2  yards,  at  12  dollars.  In  both  cases,  the  cost  is  24  dollars  ;  for, 
in  the  second,  you  divide  the  quantity  by  3,  but  increase  the  price 

3  times. 

5.  This  principle  is  useful  in  simplifying  multiplica- 
tion, in  some  cases,  as  will  be  seen  hereafter  (XLV.),  and 
may  also  be  employed  to  prove  multiplication.  Thus :  I 
find  594x437=259,578;  and  remarking,  that  594  can 
be  divided  by  3, 1  divide  it,  and  multiply  the  other  factor 
by  3,  which  gives  me  198  X  1311  =  259,578,  as  before. 

Questio7is. — If  the  multiplicand  is  made  a  number  of  times 
larger,  how  will  the  product  be?  Why?  Give  an  example. 
Same  questions  for  a  smaller  multiplicand;  also  for  the  multi- 
plier. Is  the  result  the  same  if  you  multiply  the  product  in- 
stead of  the  multiplicand  ?  Ditto,  instead  of  the  multiplier  ? 
"When  should  either  be  preferred  ?  How  is  the  product  when  one 
of  the  factors  is  multiplied,  and  the  other  divided,  by  the  same 
number  ? 

LESSON  XXIV. 

SUPPLEMENT    TO   DIVISION. 

1.  Since,  in  division, 
the  dividend,  ]    correspond,    (  i^^^  product, 

divisor,     >  respectively,  <         multiplicand, 
and  quotient  )  ^°  (  and  multiplier 

of  multiplication,  it  may  be  anticipated  that  the  princi- 
ples enunciated  above  in  multiplication,  have  correspond- 
ing ones  in  division. 

PROPOSITION    I. 

The  divisor  remaining  the  same,  if  the  dividend  he 
made  a  certain  number  of  times  larger  or  smaller  (that 
is,  if  it  be  multiplied  or  divided  by  a  certain  number)  the 
quotient  will  be  made  the  same  number  of  times  larger 
or  smaller. 

For,  a  larger  dividend  must  contain  the  same  divisor  a 
greater  number  of  times,  and  a  smaller  dividend  contain 
it  a  less  number. 


80  LESSON    XXIV. 

Also,  the  dividend  being  the  product  of  the  divisor  by 
the  quotient,  must  contain  the  divisor  as  nnany  times  as 
there  are  units  in  the  quotient  (XI.,  3) ;  consequently, 
the  number  of  units  of  the  quotient  must  be  increased  or 
decreased  as  many  times  as  the  dividend,  in  order  that 
the  new  dividend  may  still  be  produced  by  multiplying 
the  same  divisor. 

Example, — A  basket  containing  12  apples,  divided  .among  3 
boys,  gives  each  4  apples  as  a  quotient.  5  such  baskets  would 
be  5  dividends;  the  division  of  w^hich  would  give  5  times  as 
many  apples  to  each  boy — that  is,  5  times  the  first  quotient  4,  or 
20  apples. 

Again,  one-half  the  number  of  apples  gives  each  only 
one-half  his  former  share  ;  that  is,  2  instead  of  4  apples. 

2.  Hence,  we  may  multiply  or  divide  the  quotient^  in- 
stead of  the  dividend^  when  it  is  more  convenient, 

PROPOSITION   II. 

3.  The  dividend  remaining  the  same^  if  the  divisor  he 
made  a  certain  number  of  times  larger^  the  quotient  will 
be  made  as  many  times  smaller ;  and  if  it  be  made 
smaller,  the  quotient  will  be  made  as  many  times  larger. 

For  a  larger  divisor  must  be  multiplied  by  a  smaller 
quotient,  and  a  smaller  divisor  by  a  larger  quotient,  to 
produce  the  same  dividend. 

In  other  words,  the  same  dividend  contains  a  larger 
divisor  a  less  number  of  times,  and  a  smaller  one  a  greater 
number  of  times. 

Example. — 24  dollars  will  buy  6  yards,  at  4  dollars  a  yard; 
while  they  will  buy  only  3,  at  8  dollars ;  but  will  buy  12,  at 
2  dollars. 

Hence,  instead  of  multiplying  or  dividing  the  quotient, 
we  may,  when  more  convenient,  divide  or  multiply  the 
divisor,  and  vice  versa. 

Example. — 4  pieces  of  cloth,  containing  each  25  yards,  cost 
4,200  dollars.  How  much  is  it  a  yard  ?  We  might  divide  by  4, 
first,  to  find  the  cost  of  a  piece,  and  then  by  25,  for  that  of  a 
yard;  but  it  is  simpler  to  multiply  4  by  25  first,  and  then  divide 
4200  by  the  product  100. 


MULTIPLICATION    OF    DECIMALS.  St 

PROPOSITION    III. 

4.  If  the  dividend  and  the  divisor  he  each  multiplied 
or  divided  by  the  same  number ^  the  quotient  will  not  be 
changed. 

For,  a  number  of  dividends  will  contain  the  same  num- 
ber of  divisors  as  many  times  as  a  single  dividend  contains 
a  single  divisor. 

Example, — There  is  a  bag  of  silver,  containing  200  dollars  for 
each  company  of  100  men.  The  division  will  give  each  soldier 
2  dollars.  Now,  it  matters  not  whether  the  number  of  compa- 
nies be  increased  or  diminished,  provided  the  number  of  bags  be 
increased  or  diminished  in  the  same  proportion,  and  there  conti- 
nues to  be  one  bag  to  each  compan)'-;  the  division  of  the  whole 
sum  of  money  among  all  the  soldiers  must  give  to  each  the  same 
amount,  2  dollars,  as  the  division  of  a  single  bag  by  a  single 
company. 

Otherwise,  the  dividend  is  the  product  of  the  divisor 
by  the  quotient ;  and,  consequently,  the  multiplication 
of  the  same  quotient  by  a  divisor  a  number  of  times 
larger  or  smaller,  must  produce  a  dividend  as  many  times 
larger  or  smaller. 

Questiojis. — How  is  the  quotient,  when  the  dividend  is  in- 
creased a  number  of  times  ?  How,  when  it  is  decreased  i  Why  ? 
To  w^hat  do  the  dividend,  divisor,  and  quotient  correspond  in 
multiplication?  May  the  quotient  be  multiplied  or  divided,  in- 
stead of  the  dividend  ?  Why  ?  When  is  it  preferable  ?  If  the 
divisor  be  increased,  how  is  the  quotient  ?  Why  ?  If  it  be  de- 
creased? Why?  Give  examples.  May  the  divisor  be  multi- 
plied, instead  of  dividing  the  dividend  ?  May  it  be  divided 
instead  of  multiplying  the  dividend  ?  Why  ?  If  the  dividend 
and  divisor  be  either  divided  or  multiplied  by  the  same  number, 
how  is  the  quotient  ?    Why  ?     Give  examples. 

LESSON  XXV. 
MULTIPLICATION  OF  DECIMALS. 

1.  Let  it  be  proposed,  for  example,  to  operation. 

multiply 5.407 

by         -  2.54 

In  the  first  place,  let  us  consider,  for  a  qi^oq 

moment,   the    multiplier    as    the   whole  070^5 

number,  254;  it  is  clear  that  the  multi-  10814, 
plicand,  5.407,  being  (XXL,  1  and  2)  the 

same  thing  as  5407  thousandths,  the  re-  13.73378 


82  LESSON   XXV. 

petition  of  this  number  254  times,  would  give  a  product 
of  the  same  nature  (XL,  7) ;  that  is,  1373378  thousandths, 
which,  expressed  decimally,  is 

1373.378. 

2.  So  that,'  in  tlie  multiplication  of  a  decimal  number 
by  a  whole  number,  the  product  contains  as  many  decimal 
places  as  the  multiplicand. 

3.  But,  in  considering  the  multiplier  2.54  as  a  whole 
number,  we  have  increased  it  one  hundred  times ;  or,  in 
other  words,  ten  times  for  each  decimal  place  it  contains 
(XXI.,  6).     The  product 

1,373.378 

is,  therefore,  one  hundred  times  too  large ;  and  it  must 
now  be  divided  by  100,  to  reduce  it  to  its  true  value, 

13.73378. 

This  IS  done  by  removing  the  units'  point  (XXI.,  6)  two 
additional  places  to  the  left,  which  makes  in  all  as  many 
places  as  there  were  decimals  in  both  the  multiplicand 
and  multiplier.     Hence : 

I.  Midtiply  decimal  numbers  as  if  there  was  no  deci- 
mal point ;  andy  in  the  product,  point  off  as  many  figures 
as  there  are  decimals  in  both  the  multiplicand  and  the 
multiplier. 

n.  If  there  be  not  so  many  figures  in  the  product,  the 
deficiency  is  supplied  by  zeros,  between  the  product  and 
decimal  point. 

As  in  the  following  example  :  operation. 

Let  it  be  proposed  to  multiply  -  0.03054. 

by  -  0.023 

If  we  multiply  without   regard  to  the  9162 

units'  point,  we  get  the  product,  6108 

70242,  ^  0.00070242 

which  is  much  too  large ;  for,  it  should, 
in  the  first  place,  contain  five  decimals, 
if  the  multiplier  were  the  whole  number,  23 :  and,  in 


MULTIPLICATION   OF  DECIMALS.  83 

the  next,  since  the  multiplier  is  one  thousand  times 
smaller,  we  must  cut  off  three  additional  decimals  in  the 
product;  in  all,  eight  decimal  figures ,'  namely, ^ve  on 
account  of  the  multiplicand,  and  three  on  account  of  the 
multiplier. 

The  last  figure  must,  therefore,  be  of  the  eighth  order 
of  decimals.  We  move  it  to  that  place  by  prefixing  the 
requisite  number  of  zeros  between  the  product  and  deci- 
mal point,  as  shown  above. 

REMARK. 

4.  The  multiplication  of  decimal  fractions,  unconnected  with 
whole  numbers,  presents  an  apparent  contradiction,  which  is 
calculated  to  embarrass  beginners ;  namely,  that  the  product  of 
decimals  is  smaller  than  either  factor. 

Thus,  for  instance,  when  we  multiply        -        -        0.3 

by        -        0.02 


we  get  for  the  product        -        -         -         -         0.006 

which  is  of  the  third  descending  order,  and  consequently  smaller 
than  the  multiplicand,  which  is  of  the  first,  and  the  multiplier, 
which  is  of  the  second.  Hence,  the  pupil  is  at  a  loss  to  under- 
stand whether  the  operation  is  a  multiplication  or  a  division. 

The  fact  is,  that  there  is  both  a  division  and  multiplication  in 
it.  In  order  to  understand  this,  it  must  be  recollected  that  the 
object  of  Tnidti'plicalion  is,  knowing  the  value  corres'ponding  to 
one  unit,  to  find  the  value  corres'ponding  to  several ;  as  in  this 
example  : 

One  yard  of  cloth  costs  6  dollars.  What  w^ill  4  yards  cost  ? 
The  answer  is  evidently  obtained  ly  repeating  the  ■multiplicand^ 
6  dollars,  as  many  times  as  there  are  units  in  the  multiplier,  4 ; 
because  the  multiplicand  is  equal  in  value  to  each  one  of  these 
units. 

But  if  the  question  were  : 

One  (yard)  costs  6  (dollars),  what  will  4  hundredths  (of  a  yard) 
cost  ? 

The  case  would  be  different;  for,  then,  the  multiplicand, 
6  dollars,  is  no  longer  given  as  the  value  of  one  of  the  units  in 
which  the  multiplier  is  expressed,  which  are  hundredths,  but  it 
is  the  value  of  a  whole  unit  (a  yard)  one  hundred  times  larger. 
Consequently,  were  we  to  repeat  this  multiplicand,  6  dollars,  as 
many  times  as  the  number,  4,  of  units  in  the  multiplier,  we  would 
get  a  product  one  hundred  times  too  large. 


84  LESSON   XXV. 

A  preparatory  operation  must,  therefore,  be  introduced  before 
the  multiplication  can  be  correctly  performed. 

This  operation  is  a  division.  It  consists  in  reducing  the  mul- 
tiplicand, so  that  its  value  may  correspond  to  the  smaller  frac- 
tional unit,  one.  hundredth,  in  the  multiplier ;  that  is,  since  one 
whole  unit  (yard)  is  equivalent  to  the  given  tmdtipli.ca'nd  (6  dol- 
lars), the  one  hundredth  jpart  of  the  whole  unit  must  be  equivalent 
to  the  one  hundredth  'part  of  the  given  multiplicand,  6. 

Which,  by  the  principle  of  decimal  numeration,  is  0.06 
(XXL,  6);  and,  this  being  the  true  amount  equivalent  to  the 
unit,  one  hnndredth  of  the  multiplier  may  now,  with  propriety, 
be  considered  as  the  true  multiplicand  ;  and,  as  such,  be  repeated 
as  many  times  as  there  are  units  in  the  multiplier  (in  this  case, 
four  times.  The  operation  is  thus  reduced  to  a  genuine 
multiplication  of       -  -  -  -  -  0.06 

by  -  4 

and  the  true  product  is  -  -  -  -  0.24 

which  is  now,  as  usual,  larger  than  its  true  multiplicand. 

This  product,  24  hundredths  of  a  dollar,  is  the  same  that  would 
have  been  obtained  by  the  application  of  the  above  rule  of  deci- 
mal multiplication. 

It  will  be  seen,  therefore,  that  the  whole  difficulty  arises  from 
the  fact,  which  is  not  generally  noticed,  that  this  is  a  compound 
operation ;  the  first  part  of  which  is  in  fact  a  division,  to  reduce 
down  the  given  multiplicand  from  its  value,  corresponding  to  a 
large  unit,  to  the  standard  of  the  smaller  units  of  which  the  mul- 
tiplier is  composed. 

The  second,  which  is  the  true  multiplication,  to  repeat  as  usual 
this  altered  multiplicand  as  many  times  as  there  are  units  in  its 
multiplier. 

In  practice,  the  multiplication  and  the  preparatory  division  are 
blended  in  on*^  operation ;  and  the  proper  reduction,  instead  of 
being  previou.sly  made  in  the  multiplicand,  is  made  in  the  pro- 
duct, because  it  is  simpler  and  amounts  to  the  same  thing. 

5.  The  above  explanation  shows  how  we  must  understand  the 
principle  that  the  multiplication,  by  descending  units  10,  100, 
1000,  ^c,  times  smaller,  gives  products  decreasing  also  10,  100, 
1000,  ifc,  times,  like  the  orders  multiplied  by. 

It  is  an  analogy,  in  regard  to  the  decreasing  of  products  by 
descending  units,  corresponding  to  the  increase  by  ascending 
orders ;  and  which  implies  that  the  given  multiplicand,  in  regard 
to  whole  units,  is  first  reduced  down  to  correspond  to  one  of  the 
fractional  units. 

N.  B. — These  remarks,  applied  to  our  first  example,      35.407 
multiplied  by        .         .         .     12.54 

=  444.00378 


MULTIPLICATION    OF    DECIMALS. 


s§ 


may  also  serve  to  explain,  in  another  way,  the  principle  for  cutting 
pfF  decimals.  For,  the  multiplicand,  35.407,  is  here  given  as  equi- 
valent to  one  whole  unit  of  the  integral  part,  12,  of  the  multiplier, 
which  is  a  mixed  number. 

Consequently,  in  order  to  multiply  by  1254  as  a  single  number, 
we  must  determine  first  that  reduced  multiplicand  which  corre- 
sponds to  one  of  its  last  units  (viz  :  hundredths^. 

This  is  done  by  cutting  off  as  many  additional  decimals  in  the 
multiplicand  as  there  are  in  the  multiplier,  which  here  makes  it 
.35407.  By  this  preparatory  division,  the  multiplier  becomes 
a  single  abstract  number ;  and  the  new  multiplicand  containing 
as  many  decimals  as  there  were  in  both  multiplicand  and  multi- 
plier, we  must  cut  off  the  same  number  in  the  product,  since  it 
is  of  the  same  nature  as  the  multiplicand. 

Qtiestions. — What  is  the  rule  for  the  multiplication  of  decimals  ? 
How  many  decimals  do  you  cut  off?  Why?  What  is  done 
when  the  number  of  figures  in  the  product  is  not  sufficient  for 
decimals  ?  Give  examples.  When  is  the  product  of  two  decimal 
numbers  smaller  than  either  factor  ?  Is  the  operation  in  that 
case  truly  a  multiplication  ?  Is  it  a  single  operation  ?  How  can 
the  statement  be  rectified?  How  is  the  product  of  a  number  by 
a  multiplier  composed  of  descending  units  ?  Explain  the  cutting 
off  of  decimals  by  the  previous  transformation  of  the  multiplicand. 

EXERCISES. 

In  the  following  exercises,  the  decimal  point  is  purposely 
omitted;  the  fixing  of  its  place  being  the  only  difference  between 
common  and  decimal  multiplication. 

1.  12.34  X  5.6=  69104. 

2.  0.123  X  0.45=  5535. 

3.  0.321096  X  .2465=  7150164. 

4.  79.347X23.15=183688305. 

5.  0.0253  X  345=  87285. 

6.  0.0010001  X  .001=  10001. 

7.  724.623  X  5.56=  402890388. 

8.  11785  X  .027=318195. 

It  is  unnecessary  to  multiply  examples  here.  The  teacher 
may  use  any  of  the  exercises  previously  given  in  multiplication 
and  division,  by  introducing  decimal  points  in  any  place  he  may 
choose. 

8 


86  •  LESSON   XXVI. 

LESSON  XXVI. 
DIVISION  OF  DECIMALS, 

CASE    1. 

1.  The  divisor  being  a  whole  number. 

If  we  had  to  divide  a  decimal  by  a  whole  number,  as 

80.64.  by  9, 

the  quotient  would  be  8.96  ;  for,  it  is  clear  that  the  units 
of  the  quotient  being  always,  in  such  a  case  (XV.,  3),  of 
the  same  nature  as  those  of  the  dividend,  which  here  is 
8064  hundredths^  the  quotient  must  be  a  number  of  hun- 
dredths:  that  is,  896  hundredths  ;  or,  decimally,  8.96. 

Hence,  the  quotient  contains  as  many  decimal  'places 
as  the  dividend,  when  the  divisor  is  a  whole  number. 

CASE    II. 

2.  When  the  divisor  contains  decimals. 

In  this  case,  all  we  have  to  do  is  to  make  the  divisor  a 
whole  number,  and  to  divide,  as  in  the  preceding  case. 
Let  us  take  the  annexed 


FIRST   EXAMPLE. 


OPERATION. 


36/8.64 
326 
384 


4/8 


We  make,  at  once,  the  divisor  a 
whole  number  by  striking  off  its 
decimal  point. 

But,  in  doing  this,  we  increase 
the  divisor  ten  times  for  each  deci- 
mal place  it  contains  (XXL,  6); 
and  consequently  the  quotient  would  be  made  the  same 
number  of  times  smaller,  if  the  dividend  remained  un- 
changed (XXIV.,  3). 

In  order,  therefore,  that  the  quotient  may  not  be 
altered,  we  must  increase  the  dividend  as  many  times  aa 
the  divisor. 

This  we  do  by  removing  its  units'  point  as  many  places 
*o  the  right  as  there  were  places  in  the  divisor. 


DIVISION   OF   DECIMALS.  87 

Then  the  operation  is  performed  as  usual ;  and  the 
units'  point  is  placed  in  the  quotient  when  the  operation 
reaches  it  in  the  new  dividend;  or  else,  we  may,  after 
having  found  the  whole  quotient, 

768 
fix  its  character  hy  cutting  off  the  same  number  of  decimal 
places  as  were  left  in  the  dividend,  (XV.,  3). 

GENERAL  RULE. 

3.  I.  Mahc  the  divisor  a  whole  numher  hy  striJcing  off 
its  units*  point,  and  remove  the  units'  point  of  the  dividend 
as  many  places  to  the  right  as  there  were  decimal  figures 
in  the  divisor,  annexing  ciphers,  if  necessary . 

II.  Divide,  then,  as  in  whole  numbers,  and  place  the 
units'  point  after  the  quoiientfigure  of  units. 

III.  If  the  first  partial  dividend  is  decimal,  set  down  a 
zero  for  units  and  one  for  each  vacant  order  after  it. 

N.  B. — This  rule  embraces  all  the  cases  usually  separated  in 
Arithmetics,  and  is  more  easily  understood  and  applied  by  begin- 
ners. 

No  confusion  can  result  from  the  two  decimal  points  in  the 
dividend,  because  it  is  always  that  on  the  right  which  determines 
the  order  of  the  quotient. 


1.92 


OPERATION. 

368.64 
1766 
384 


192 


SECOND  EXAMPLE. 

4.  When  the  number  of  decimals 
is  the  same  on  both  sides,  though 
the  rule  would  apply,  it  is  evident 
that  no  preparation  is  necessary ; 
and  we  may  divide  at  once,  with- 
out regarding  the  decimal  point ; 
for,  this  amounts  to  increasing  equally  the  dividend  and 
divisor.  Then  the  quotient,  to  the  end  of  the  dividend, 
is  a  whole  number. 

THIRD    EXAMPLE. 

5.  TVhen  the  number  of  decimals  is  less  in  the  divi' 


^5  LESSON   XXVI. 

dend,  the  removal  of  the  units'  point  requires  the  addi- 
tion of  some  Osj  it  is  the  case  mentioned  in  the  rule. 

Let  it  be  proposed  to  divide  3686.4  by  1.152 

OPERATION. 

3686/40011/152 
..2304     1-^^ 
...      1 

FOURTH  EXAMPLE. 

.025  by  0.5 
/0.25|0;5 

••  loToH" 

FIFTH    EXAMPLE. 

0.25  by  0.00005 


Which  is  changed  into  0^25000  by 


o;oooo5 


5000. 


6.  In  these  two  examples,  and  especially  in  the  last, 
we  see  that  the  quotient  is  larger  than  the  dividend ;  the 
last  being  5,000  whole  units,  while  the  dividend  was 
only  the  small  fraction,  25  hvndredths. 

This  result  is  more  readily  understood  than  the  de- 
crease of  the  product  in  the  multiplication  of  decimal 
fractions.  For,  however  small  a  quantity  may  be,  there 
may  always  be  conceived  a  much  smaller  one,  which  will 
be  contained  in  it  a  great  number  of  times. 

Questions. — Give  the  rule  for  dividing  decimal  fractions,  when 
the  divisor  is  a  whole  number  ;  when  it  contains  decimals.  What 
do  you  make  the  divisor,  if  you  do  away  with  its  units'  point  ? 
What  must  be  done,  after  this,  with  the  dividend  ?  Why  ?  If  the 
number  of  decimal  places  in  the  dividend  is  smaller,  what  is 
done  ?  How  many  decimals  must  there  be  in  the  quotient  ? 
When  the  number  of  decimal  places  is  the  same  in  both  dividend 
and  divisor,  how  do  you  divide  ?  Is  the  quotient  ever  larger 
than  the  dividend  ?     Explain  it. 

EXERCISES. 
The  units'  points  and  annexed  ciphers,  which  constitute  the 


OF    DECIMAL    REMAINDERS.  89 

principal   difficulty  of  the   following  exercises,  are  purposely 
omitted  in  the  answers. 


1.482.2635-^32.1509=15. 

2.  559.650 -f- 2.5  =  22386. 

3.  234.70525 -f.  3.653  =6425. 

4.  0.7854-^-1.4=561. 

5.  0.87275718-1-0.162  =  538739. 

6.  0.00070242-^0.023=3054. 


7.  116.0435-1-0.29  =  40015. 

8.  7254.06-1-0.0758  =  957. 

9.  118.5-1-0.125=948. 

10.  0.5822-1-142  =  41. 

11.  0.00102048^31.89  =  32. 

12.  0.99-^0.00225  =  44. 


What  is  one  tenth  divided  by  one  tenth  ?  One  tenth  by  one 
hundredth  i  One  hundredth  by  one  tenth  ?  By  one  thousandth  ? 
By  one  hundredth  ?  Ten  units  by  one  hundredth  ?  One  hundredth 
by  ten  units  ?  One  thousandth  by  one  thousandth  ?  One  tenth 
by  thousand  units  ?  &c.  How  often  are  ten  units  contained  in 
one  tenth  ?  One  tenth  in  ten  units  I  One  hundredth  in  one  unit  ? 
Fifteen  units  in  fifteen  hundredths  ?  Two  hundred  units  in  two 
hundredths?     Sixty-five  tenths  in  sixty-five  units?  &c.  &c. 


LESSON  XXVII. 
OF  DECIMAL  REMAINDERS. 

1.  When  a  division  of  decimals  hy  a  whole  numher 
leaves  a  remainder,  it  is  a  number  of  units  of  the  same 
order  as  the  last  units  of  the  dividend. 

Set  it  over  the  divisor^  and  join  it  to  the  quotient^  as 
in  simple  division. 


OPERATION. 


il45 


EXAMPLE. 

Let  it  be  proposed  to  divide  .3.723944  by  0.145 

After  having  prepared  the 
operation,  as  usual  (XXVL,  3), 
we  see  that  the  altered  dividend 
is  of  the  order  of  thousandths. 
The  remainder  is  consequently 
54  thousandths,  which  are  yet 
to  be  divided  by  145.  Therefore, 
the  expression,  -^-^j  of  this  divi- 
sion, annexed  to  the  quotient, 
completes  it. 
8* 


3/723.944 
•823 
989 
1194 
•344 
54 


25.682//y 


90 


LESSON    XXVII. 


25.6823724-//^ 

or 
25.6823724  + 


EXTENSION  OF  THE  QUOTIENT. 

2.  Had  we  stopped  at  the  whole  number,  25,  for  the 
quotient,  the  remainder  would  have  been  98  whole  units, 
and  the  quotient  itself  would  not  have  been  so  exact  avS 
that  we  have  obtained  by  changing  the  remainder  into 
tenths,  then  hundredths,  &c.,  and  getting  corresponding 
figures  in  the  quotient. 

3.  We  may,  in  the  same  way,  get  a  still  greater  ap- 
proximation by  changing  the  final  remainder  into  smaller 
orders,  and  dividing  these  also  by  the  divisor.  Thus  the 
quotient  may  be  extended  to  orders  as  small  as  may  be 
desired. 

For  this  purpose,  one  operation. 

cipher  is  added  to  the    3/723.944  /145 

remainder,      54     thou-      823 
sandths,  which    makes         989 
it  540  tenths  of  thou-         1194 
sandths;    and    thus    3  344 

tenths    of   thousandths  540 

are  obtained  in  the  quo-  1050 

tient.  .  350 

In  the  same  way,  to  600 

the  new  remainder,  105,  20 

add  a  new  cipher,  and 

you  get  7  hundredths  of  thousandths  in  the  quotient,  with 
a  remainder,  35.  The  addition  of  a  new  zero  to  this, 
makes  it  350  miUionths,  which,  divided  by  145,  gives 
2  millionths  in  the  quotient.  The  next  figure  is  4  tenths 
of  millionths. 

If,  from  the  nature  of  the  operation,  the  approximation 
is  deemed  sufiicient,  we  may  stop  here. 

4.  N.  B. — To  give  the  complete  quotient,  we  have  annexed  to  it 
the  indicated  division,  y:fy,  of  the  remainder;  but,  in  practice, 
the  desired  approximation  being  obtained  by  the  diminutive  size 
of  the  last  units,  the  remainder  is  generally  neglected,  and  the 
quotient  given  without  this  small  addition.  Sometimes,  in  that 
case,  the  sign  +  is  added,  as  shown  above,  to  indicate  that  the 
quotient  is  not  exact,  and  might  be  extended. 

5.  It  is  better^,  when  the  requisite  degree  of  ap^roxi- 


DIVISION    OF    DECIMALS.  91 

wation  is  fixed  heforeliand^  to  add,  at  once,  to  the  dim- 
dend  as  many  zeros  as  are  necessary  to  make  the  last 
decimal  place  of  the  dividend  of  the  desired  order. 

Wishing,  in  the  present  instance,  to  extend  the  quo- 
tient to  tenths  of  millionths,  seven  decimal  places  are 
requisite  ;  and  we  may  add  at  once  four  zeros  to  complete 
the  number  of  decimals  and  prepare  the  operation  thus  : 


3723.9440000 


145 


This  arrangement  secures  room  for  the  operation  3  a 
thing  which  should  always  be  attended  to  for  the  sake  of 
neatness  and  accuracy. 

6.  The  dividend  being  now  a  number  of  tenths  of 
millionths,  the  quotient  is,  of  course,  of  the  same  order ; 
and  the  result  is  said  to  be  obtained  true,  or  carried  to, 
icithin  one  tenth  of  millionth,  or  to  the  seventh  decimal 
place;  because,  what  might  follow,  if  the  division  were 
extended,  is  less  than  one  tenth  of  millionth  (VL,  3).* 

7.  N.  B. — Divisions  carried  out  in  that  way  will  very  frequently 
never  end,  and  give  rise  to  infinite  decimals. 

REMARK  I. 

8.  WJien  a  number,  accompanied  by  decimals,  is  to 
be  divided  by  1,  followed  by  zeros,  such  as  10, 100,  1000, 
&c.,  remove  the  units'  point  as  many  places  to  the  left  as 
the  divisor  contains  zeros  ;  the  residt  is  the  quotient. 


EXAMPLES. 

25.37  divided  by 

((           a 

10,          is 
100,        " 
1000,      " 
10,000,  « 

2.537 
0.2537 
0.02537 
0.002537 

This  results  from  Lesson  XXI.,  6. 


*  When  the  figure  after  the  last  decimal  retained,  is  either  5  or 
over,  it  is  customary  to  add  one  unit  to  the  last  decimal,  because  the 
rejected  figure  is  nearer  to  te?i  than  to  zero.  For  instance,  in  the 
above  division,  had  we  stopped  at  the  tenths  of  thousandths,  we  should 
have  taken  for  the  quotient  25.6824  instead  "of  25.6823. 


92 


LESSON   XXVII. 


If  the  dividend  should  end  with  Os,  strike  them  out  when  they 
pass  to  the  right  of  the  decimal  point,  since  they  are  of  no  use 
there  (XXL,  8  and  9).  2500,  for  example,  divided  by  1000  is  2.5. 


REMARK  II. 


9.  If  the  divisor^  being  an  integer,  ends  with  zeros^ 
they  may  he  struck  off,  and  the  decimal  point  removed 
as  many  figures  to  the  left  in  the  dividend,  when  the 
division  may  proceed  as  usual. 

For,  in  doing  this,  you  decrease  both  equally,  which 
does  not  alter  the  quotient  (XXIV.,  4). 


Let  it  be  proposed  to  divide  996.03  by  2700. 


If  the  dividend  v^ere  an  integer, 
the  case  would  be  the  same ;  some 
of  its  last  figures  then  would  be 
made  decimals. 


OPERATION. 


9.96/03 
186 
240 
243 


2700 


0.3689 


For  example : 
into     -    -    -    - 


684      to  be  divided  by  20,  would  be  changed 
68.4     to  be  divided  by  2. 


This  is  the  remark  adverted  to  in  Lesson  XVII.,  7; 
which  I  have  seen  strangely  and  yet  frequently  misap- 
plied, because  it  is  usually  given  for  whole  numbers,  and 
omitted  in  connexion  with  decimal  fractions.  I  have 
seen  boys,  in  examples  like  this, 

15.-96738  to  be  divided  by  20, 

apply  the  usual  rule,  and  prepare  the  operation  by  cut- 
ting off  in  the  divisor  the  0,  and  in  the  dividend  the 
figure  8  5  as  here  exhibited  : 

15.9673)8  20, 

and  then  divide  by  2,  without  reflecting  that  their  object 
should  be  to  decrease  the  dividend  10  times,  as  well  as 


DIVISION    OF   DECIMALS.  93 

the  divisor,  and  that  this  can  be  done  only  by  moving 
the  units'  point  3  thus  : 

1.5/96738 1 20. 


10.   I  have   seen   also  this  case.     Having  to  divide 
15.96738  by  .200,  the  operation  was  prepared  thus: 


15.967)38 


.200, 


by  striking  off  the  two  ciphers,  and  cutting  off  two 
figures  in  the  dividend,  forgetting  that  the  two  Os  struck 
out  do  not  alter  the  divisor,  nor  does  the  separation  of 
the  decimal  figures  in  the  dividend  change  its  value. 
Here  there  would  be  in  fact  no  error  in  the  result,  but  a 
mere  awkward  evidence  of  a  misapprehension  of  princi- 
ples. 

REMARK   III. 

1 1.  When  the  dividend  and  divisor,  being  both  integers, 
end  with  Os,  an  equal  number  may  be  struck  out  of  each, 
and  then  the  operation  will  be  performed  as  usual. 

EXAMPLE. 

612400      divided  by      12000. 
is  the  same  as  6124  "      by      120. 

and  as  612.4  «      by       12. 

Since  both  numbers  are  decreased  equally  in  each  case. 

Such  a  transformation  is  not  indispensable,  but  merely 
preferable. 

N.  B. — Here,  again,  recollect  that  the  zeros  struck  off, 
if  they  come  after  a  decimal  fraction,  do  not  require  a 
corresponding  operation  in  the  other  number. 

For  example : 

If  612400  were  to  be  divided  by  1.2000, 
the  removal  of  the  three  Os  of  the  divisor  would  not 


y4*  LESSON   XXVII. 

affect  its  value ;  and,  therefore,  the  dividend  must  not  be 
touched,  though  we  divide  by  1.2  instead  of  1.2000. 

Questions. — Of  what  order  is  the  remainder  of  a  decimal  divi- 
sion ?  What  is  done  with  it  ?  How  may  the  quotient  be  ex- 
tended ?  How  do  you  show  that  the  division  is  not  ended  ?  If  you 
add  a  0  to  the  decimal  remainder,  what  do  you  make  it  ?  What 
are  infinite  decimals  ?  How  would  you  prepare  the  dividend  to 
obtain  a  quotient  to  within  one  hundredth  ?  One  thousandth  ? 
One  tenth  of  thousandth  ?  &c.  What  is  meant  by  this  expression, 
to  within  one  hundredth  ?  &c.  How  do  you  divide  decimal 
numbers  by  10,  100,  1000,  &c.  ?  If  the  dividend  ends  with  zeros, 
what  is  done  when  the  divisor  is  1,  followed  by  zeros  ?  If  the 
divisor,  being  an  integer,  ends  with  zeros,  how  do  you  proceed? 
If  the  decimals  of  the  divisor  end  with  zeros,  what  is  done  ?  If 
the  dividend  and  divisor,  being  whole  numbers,  end  with  ciphers, 
what  may  be  done  ? 

EXERCISES. 

The  decimal  point  and  additional  ciphers  are  purposely  omitted 
in  the  answers. 

1.  6345.923-^-54.23=  117018 -+- 

2.  27845.96  —  9.8732=28203581  + 

3.  10-4-563=  177 -+- 

4.  7-^365=  19178  + 

5.  0.00026253 -i- 0.175=  150  + 

6.  256.6  -^  0.057  ==  450175  + 

7.  1-1-0.159  =  6289308  + 

8.  0.2-^23.2  =  862068  + 

9.  Divide  .00636056  by  .86  to  within  one  millionth. 

10.  Divide  61  by  0.825  to  the  6th  place  of  decimals. 

11.  Get  the  quotient  __8_7_?|.^^  true  to  the  5th  place. 

12.  Divide  37.96416  by  0.156  to  one  hundredths. 

13.  Divide  0.59  by  79800  to  the  12th  place  of  decimals. 

14.  Divide  23  by  0.000579  to  within  one  hundredth  of  millionth. 

N.  B.— It  being  desirable  that  the  learner  should  perform 
readily  divisions,  without  any  assistance  from  answers,  the 
teacher  may  use,  also,  some  of  the  preceding  examples  in  multi- 
plication and  division,  with  proper  and  to  him  easy  modifications, 
such  as  the  introduction  of  decimal  points,  ciphers,  and  remain- 
ders, which  will  still  leave  a  key  to  the  teacher,  unknown  to  thQ 
pupil. 


VULGAR  FRACTIONS.  95 


CHAPTER  IV. 

CONTAINING   VULGAR    FRACTIONS. 

LESSON  XXVIII. 
NATURE  OF  VULGAR  FRACTIONS. 

1.  Decimal  fractions  have  already  explained  what  frac- 
tions are )  but  units  smaller  than  the  unit  of  comparison 
are  not  always  formed  by  successive  subdivisions  by  10. 

In  a  great  many  cases,  the  relation  between  the  principal 
and  the  subordinate  units  is  expressed  by  other  numbers 
than  10  or  its  multiples  100,  1000,  &c. 

For  example  :  I  want  to  measure  the  length  of  a  board.  The 
unit  of  comparison^  for  this  purpose,  is  the  foot.  But  I  find  12 
feet  and  something  over,  less  than  a  foot,  which  I  must  necessa- 
rily measure  by  some  smaller  unit,  connected  with  the  foot  by 
some  known  relation ;  as,  for  example,  the  inch,  12  of  which  are 
equal  to  one  foot.  The  inch  is  therefore,  a  fraction  of  a  foot, 
equal  to  one  twelfth  part  of  it.     It  is  a  relative  unit. 

Again :  I  want  for  a  coat  2  yards  and  a  piece  less  than  a  yard ; 
that  is,  a  fraction  of  a  yard,  the  measure  of  which  must  be 
expressed  in  smaller  units,  bearing  a  known  relation  to  the  whole 
yard.  If  four  such  pieces,  for  instance,  are  equal  to  one  yard, 
then  I  would  ask  the  merchant  for  2  yards  and  a  quarter,  or 
fourth,  which  would  indicate  the  relation  of  the  small  additional 
piece  to  the  standard  of  measure ;  the  quarter  is  a  relative  unit. 

We  see,  in  these  cases,  that  the  number  mentioned  is 
expressed,  as  was  done  in  decimals,  by  two  kinds  of 
units ;  the  one,  the  unit  of  comparison ;  the  other,  a 
smaller  relative  unit. 

It  will  be  readily  understood,  that  the  relation  of  small 
to  large  units  may  be  expressed  in  a  variety  of  ways,  as 
infinite  as  numbers  themselves ;  for,  a  single  unit  may  be 
conceived  to  be  composed  of  any  number  of  small  parts. 

2.  This  variety  of  subdivisions  of  a  unit,  requires  a 
separate  nomenclature  and  notation. 


96  'LESSON   XXVIII,   ''^ 

When  it  takes  2  "l  small  units  ^  one  half,  written  -     -     J 

one  third,       "      -     -     i. 


iQ  "      3  I  ^^  niake  up 

J>  ^Z^e  small  re-  < 
lative    unit 
or  fraction 
6  J  is  called 


«  «     5 


one  fourth  or 

one  quarter, 

one  fifth,        "      -     -     ^ 

one  sixth,       "      -     -     ^ 

&c.  '  &c.  &c. 

These  expressions  convey  to  the  mind,  in  an  abstract 
manner,  the  relation  between  the  small  units  and  the 
whole  one, 

NOTATION. 

3.  Since  the  value  or  magnitude  of  each  small  unit  is 
fixed  by  the  number  of  parts  into  which  the  principal 
unit  has  been  divided,  it  follows  that  the  notation  used  for 
division  is  correctly  applied  here  j  for, 


The  fraction 

I 

is  one,  divided  by    - 

2 

a 

i 

«         « 

3 

ii 

4 

u         a 

4 

&c. 

&c 

The  fractional  part  of  a  thing  may  contain  more  than 
one  of  the  inferior  units  into  which  the  principal  unit 
has  been  divided. 

Instead  of  one  third,  we  may  have  two  thirds. 

cc         r         r     IV.  £c        {  two  or  three  fourths 

"       of  one  fourth,  "        -!  , 

'  (      or  quarters. 

«       of  one  fifth,  «  2,  3,  or  4  fifths. 

*'       of  one  sixth,  "  2,  3,  4,  or  5  sixths, 

&c.  &c. 

The  notation  adopted  in  this  case  is  like  the  preceding. 
The  number  of  parts  is  written  above,  and  the  relation  of 
value  of  one  of  them  to  the  principal  unit  is  written 
below,  the  line  of  division. 


VULGAR    FRACTIONS.  97 


Thus :  f  for  two  thirds. 


J  for  three  fourths  or  quarters. 
f  for  four  fifths, 
f-  for  six  sevenths, 
■j^j  for  nine  twelfths,  &c. 

These  are  termed  Vulgar  Fractions,  to  distinguish 
them  from  decimal  fractions. 

definitions. 

4.  The  two  numbers  used  to  write  a  fraction,  are  its 
two  terms. 

The  upper  number  is  the  numerator,  so  called  because 
it  expresses  the  number  of  parts. 

The  lower  number  is  the  denominator,  because  it  ex- 
presses the  relative  denomination  of  the  inferior  unit, 
with  reference  to  the  principal  one. 

READING  OF  VULGAR  FRACTIONS. 

5.  Read  the  numerator  as  a  simple  number,  and  then 
the  number  in  the  denominator,  with  the  addition  of  the 
proper  termination. 

This  termination  beyond  thirds  is  generally  th,  except  with 
numbers  over  twenty  ending  with  one,  which  is  then  called  first,  as 
5 J  [one  twenty-first,) 

With  those  ending  with  2,  which  is  then  called  second,  as  g^ 
(iiine  twenty -seconds^. 

And  with  those  ending  with  3,  which  is  then  called  third,  as 
X.  {seven  forty -thirds^. 

But  \%  would  read  sixteen  twenty -ninths ;  -^  eleven  forty" 
sevenths,  &c. 

IDENTITY  OF  FRACTIONS  AND  THE  QUOTIENTS 
OF  DIVISION. 

6.  The  notation  adopted  for  fractions  is  the  same  as 
that  for  division;  because,  in  fact,  a  fraction  is  nothing 
hut  the  expression  of  the  division  of  its  numerator  by  its 
denominator* 

9  6 


98  LESSON   XXVIII. 

Let  us  take,  for  example,  the  fraction  | ;  which  means 
3  units,  each  of  which  is  the  one  eighth  part  of  the 
principal  unit ;  that  is,  3  times  -J-. 

It  will  be  easily  understood  that  3  times  |  is  the  same 
thing  as  3  divided  by  8  :  for,  let  us  suppose  that  we  have 
3  things,  say  3  apples,  and  wish  to  divide  them  equally 
among  8  boys.  Having  divided  each  into  8  parts,  it 
will  make  no  difference  whether  we  give  to  a  boy  -J  of 
each  apple,  making  in  all,  for  his  share,  3  times  ^ ;  or 
else  3  pieces  of  the  same  apple,  which  are  |  of  it. 

7.  From  this  it  also  appears  that  3  is  equal  to  8  times  |. 
For,  8  eighths  of  anything,  is  the  thing  itself. 

8.  Hence  we  see  here  only  a  change  of  names,  and 
that  a  fraction  is  nothing  but  the  expression  of  a  division, 
in  which 


The  Dividend 

is  called         the  Numerator  ; 

the  Divisor 

"                the  Denominator  ; 

A  the  Quotient 

"                the  Fraction  itself. 

So  that    J    is  the 

quotient  of     3 

divided 

by    4. 

3 

«             of    3 

C( 

by    8. 

1 

«             of    4 

C( 

by    5. 

A 

«             of    5 

cc 

by  12. 

^ 

«            of  16 

a 

by  99. 

&C 

&C. 

This  is  the  true  character  of  fractions ;  and  much  of 
the  obscurity  which  learners  find  in  this  subject,  arises 
from  this  difference  of  names  to  express  the  same  thing. 
This  is  not  the  last  example  of  the  imperfection  of  arith- 
metic in  this  respect. 

Hence,  when  we  placed  the  remainder  of  division 
over  the  divisor,  we  used  a  fraction  in  reality  to  indicate 
a  small  addition  to  the  quotient  necessary  to  complete  its 
exact  value. 

If  we  divide,  for  example,  19  by  4,  the  exact  quotient  is  4| ; 
that  is,  4  whole  units  and  3  smaller  relative  units,  equal  each  to 
only  one  fourth  part  of  the  principal  one. 


VULGAR    FRACTIONS.  09 

9.  We  may  notice  here  that  decimal  fractions  may  be 
put  under  the  form  of  common  fractions,  and  that 

0.1         is  the  same  thing  as  i  , 

0.02                   «                   as  _2_. 

0.125                «                   as  J^2_5_. 

0.0003              «                  as  T^3_. 

10,000* 

&c.  &c. 

DEFINITIONS. 

10.  The  word  Fraction,  in  its  most  extended  sense,  is 
applied  to  all  divisions,  which,  when  merely  indicated, 
are  called  fractional  expressions'oFJractional  numbers.    { 

11.  The  different  kinds  of  fractional  expressions  are 
distinguished  sometimes  by  the  following  names,  which 
are  not,  however,  of  much  importance.  A  iiro'per  frac" 
tion  is  that  whose  numerator  is  smaller  than  its  denomi- 
nator j  as,  i,  f,  I,  j\,  &c. 

An  improper  fraction  is  that  whose  numerator  is 
equal  to,  or  larger  than,  its  denominator  j  such  as,  f,  |-, 
y/,  &c.  It  is  another  name  for  a  common  division, 
merely  indicated  by  the  conventional  sign  of  the  opera- 
tion. 

A  mixed  number  is  composed  of  a  whole  number  and 
a  fraction  joined  together;  such  as,  2^,  4i,  7|,  &c. :  also, 
4.07,  6.25,  &c.  It  is  nothing  but  the  usual  form  of  the 
complete  quotient  of  a  division  -,  as,  for  example,  -J  =  2^. 

A  simple  fraction  is  a  single  fraction,  like  ^,  |,  J,  &c. 

A  compound  fraction  is  a  fraction  of  fractions  ;  or,  in 
other  words,  several  fractions  connected,  with  the  word 
of  between  them  ;  such  as,  ^  of  ^ ;  f  of  |-  of  f . 

A  complex  fraction  is  that  of  which  one  or  both  terms 
are  mixed  numbers ;  as, 

21        31         2 


&c. 
It  will  be  seen  that  a  proper  fraction  is  less  than  1, 


3     6i'   ^' 


100  LESSON  XXIX. 

and  that  an  improper  fraction  is  equal  to  1,  when  its  two 
terms  are  the  same,  and  more  than  1,  when  the  nume- 
rator is  the  larger. 

Questions. — What  is  the  origin  of  fractions  ?  What  is  the  dif- 
ference between  the  unit  of  a  fraction  and  the  unit  of  comparison  ? 
What  is  a  relative  unit  ?  How  many  kinds  of  vulgar  fractions 
are  there  ?  How  do  you  write  a  fraction  ?  What  is  the  nume- 
rator ?  The  denominator  ?  Both  together  ?  What  are  vulgar 
fractions  ?  How  do  you  read  a  fraction  ?  What  do  you  call  one 
part  of  a  thing  divided  into  2  ?  3  ?  4  ?  5  ?  6  ?  &c.,  parts  ?  An 
apple  is  cut  into  six  parts  :  how  do  you  call  2  ;  3 ;  4  ;  5,  of  them  ? 
How  many  fifths  in  2  ?  in  3?  in  4?  in  5  apples?  How  many 
thirds  in  two  apples  and  one  third ;  how  many  fourths  in  three 
and  three  quarters  ?  &c. 

How  do  fractions  compare  with  division  ?  Show  that  a  frac- 
tion is  the  quotient  of  its  numerator  by  its  denominator.  Give 
examples.  What,  in  a  fraction,  corresponds  with  the  dividend  ? 
Divisor  ?  Quotient  ?  Can  decimal  fractions  be  written  as  vul- 
gar fractions  ?  What  is  the  quotient  of  6  by  7  ?  of  11  by  12  ?  of 
19  by  20?     Write  0.06;  0.175;  0.0061,  &c.,  as  vulgar  fractions. 

What  are  fractional  numbers  ?  fractional  expressions  ?  proper, 
improper,  simple,  compound,  complex  fractions  ?  What  is  a 
mixed  number  ?  Give  examples.  What  are  fractions  of  frac- 
tions ?  What  fractions  are  smaller  than  1  ?  equal  to  1  ?  larger 
than  1  ?  How  many  times  is  t.  smaller  than  1  ?  |.  than  3  ?  _5_ 
than  5?  &c.  ^  ^  ^^ 

LESSON  XXIX. 
FUNDAMENTAL  PROPOSITIONS. 

1.  It  being  well  understood  that  a  fraction  is  the  expres- 
sion of  a  division,  and  its  value  that  of  the  quotient,  we 
may  repeat  and  apply  to  fractions  the  propositions  rela- 
tive to  division,  given  in  Lesson  XXIV.,  by  merely 
changing  the  w^ords : 

Dividend  into  Numerator ; 

Divisor  "  Denominator  j 

Quotient  *'  Fraction. 

PROPOSITION    I. 

2.  The  denominator  remaining  the  same,  if  the  nume- 
rator  of  a  fraction  he  mnltiplied,  or  divided,  hy  a  certain 


VULGAR    IRACTIONS.  101 

numhc7%  the  fraction  itself  is  made  as  many  times  largery 
or  smaller,  as  there  are  units  in  the  number  we  multiply^ 
or  divide,  by. 

Because  the  numerator  is  another  name  for  dividend  ; 
and  we  know  that,  in  multiplyinjj  the  dividend,  we  in- 
crease the  quotient,  and,  in  divid5ng,,ideic:''easc  it  (XXIVl,!'). 

EXAl^PLES^    .       :  .'  ;'>\,  -^'o  :;.  *  ^ 

If  we  multiply  by  2,  the  numerator  of  the  fraction  |-,  we  get 
6    which  is  double  1. 

4'  4 

Again :  twice  3  quarters  of  a  yard  is  6  quarters. 

If,  in  the  fraction  y^,  we  divide  9  by  3,  we  get  j\-,  which  are 
3  small  units,  instead  of  9. 

In  decimals,  -yq  would  be  0.9 ;  which,  divided  by  3,  is  0.3  = 
_3_;  a  result  identical  with  the  preceding. 

PROPOSITION    II. 

3.  The  numerator  remaining  the  same,  if  the  denomi- 
nator be  multiplied,  or  divided,  by  a  certain  number ,  the 
fraction  itself  is  made  as  many  times  smaller ,  or  larger, 
as  there  are  units  in  the  number  we  multiply,  or  divide, 
by. 

For  the  denominator  is  another  name  for  the  divisor  ; 
and  we  know  (XXIV.,  3)  that,  in  multiplying  the  divisor, 
we  decrease  the  quotient ;  and,  in  dividing,  increase  it. 

EXAMPLES. 

If,  in  the  fraction  1^  we  multiply  the  denominator  by  2,  we 
get  |.,  which  is  twice  as  small;  for,  each  eighth  is  one  half 
of  one  quarter,  and  three  of  them  one  half  of  three  quarters. 

Again :  a  quarter  of  an  apple  cut  in  two,  makes  two  eighths  ; 
and  if  I  give  a  boy  three  of  these  'pieces,  that  is,  3  eighths,  I  give 
him  half  as  much  as  if  he  had  3  quarters. 

If,  in  -j-^Q-,  we  divide  10  by  5,  we  get  -|,  which  is  5  times  as 
large ;  since  each  half  is  worth  J-,  and  1  is  consequently  ±5 
Decimally,  4.5  is  5  times  0.9. 

One  half  of  an  apple,  cut  into  5  pieces,  would  make  5  tenths. 
tI^'  by  the  decimal  notation,  would  be  0.09 ;  atid,  if  we  divide 
9* 


102  LESSON   XXIX. 

the  denominator,  100,  by  10,  we  get  _^,  or,  decimally,  0.9,  which 
is  ten  times  as  much. 

4.  From  the  two  foregoing  propositions,  we  conclude 
that: 

I.  A  fraction 'rAa^^e  muliiplied  by  a  given  number 
in  two  differehi  ways  ;  either  by  multiplying  its  numera- 
tor: or  by  dividing  its  denominator, 

TJv  A  fraction  may  be  'divided  by  a  given  number, 
also,  in  two  different  ivays;  either  by  dividing  its  nume- 
rator or  by  multiplying  its  denominator. 

The  nature  of  the  question  will  determine,  in  each 
case,  which  of  the  two  methods  is  to  be  chosen.  In 
general,  however,  the  method  by  division,  when  it  is 
practicable,  is  preferable,  because  it  reduces  the  size  of 
the  numbers. 

EXAMPLES. 

Let  it  be  proposed  to  multiply  the  fraction  j^^  by  3. 
The  only  way  here  would  be  to  multiply  its  numerator  j 
and  the  result  would  be  H-. 

But,  if  it  were  to  be  multiplied  by  4,  since  the  deno- 
minator can  be  divided  by  4,  division  should  be  preferred, 
and  the  result  would  be  f ;  which  is  simpler  than  -^  °-. 

Let  it  now  be  asked  to  divide  |-  by  5  :  there  would  be 
no  other  way  than  to  multiply  the  denominator,  7,  by  5, 
since  the  numerator  cannot  be  divided.  Thus,  the  result 
would  be  3?y. 

But,  if  the  fraction  were  to  be  divided  by  3,  then  the 
operation  by  dividing  the  numerator,  would  be  better  5 
since  it  gives  f ,  a  simpler  result  than  2^5  which  the  other 
method  would  produce. 

Sometimes  both  methods  are  combined ;  of  which  we 
shall  find  numerous  examples  in  the  sequel. 

5.  As  a  consequence  of  the  above  rule,  we  change  a 
fraction  into  a  whole  number,  equal  to  its  numerator,  by 
merely  striking  out  its  denominator. 

For  example,  in  ^,  if  we  divide  the  denominator  by  8,  we  get  ^, 


VULGAR    FRACTIONS.  10^ 

or  3  units,  since  a  number  divided  by  unity  is  equal  to  itself; 
and,  in  fact,  we  know  that  3  is  equal  to  8  times  1  (XXVIII. ,  7) ; 
for,  3  being  the  dividend,  8  its  divisor,  and  ^  the  quotient,  the 
product  of  8  by  3  must  give  the  dividend  again. 

Questions. — What  effect  is  produced  on  the  value  of  a  fraction 
by  multiplying  its  numerator  /  By  dividing  the  same  ?  By 
multiplying  its  denominator?  By  dividing  the  same?  Why? 
Give  examples.  Repeat  each  proposition.  How  many  times  is 
6.  larger  than  2.  ?  ^4  than  -2_.  ?  |  than  _4_  ?  «.  than  ^-  ?  How 
many  times  is  2  smaller  than  ii  ?  ^  than  |.|-  ?  7  than 
7?  3  than  i.?  &c.  In  how  many  ways  can  a  fraction  be 
multiplied  by  a  given  number?  In  how  many  ways  divided? 
Which  of  the  methods  is  generally  preferred,  when  practicable  % 
Why  ?  Give  examples.  What  is  the  effect  of  striking  out  the 
denominator  of  a  fraction  ? 

EXERCISES. 
Multiply,  in  the  simplest  manner. 

By  5,     -      f;  |;  |;  |;  M;  f|;  ir^  3V;  4^;  yV  &c. 

By  7,     |;  fj   2¥'   TT'   tV'   tV'  A'  -6T'   2T'  15'  "Jl' &c. 
Divide,  in  the  simplest  way, 

-t>y  o,      -      T2'    IT'    5'    19^'     T   '     4    '    TT'    3-T'    6'    T3 J  «*C. 
Ti,,  Q  2.      9;     81.    63.    2-4.     18:27;     85.    72      o_ 

-t^y  y?      -     T'    TT'    40'     T  '    5'   "9'    Ts-'   -2  8'     T'     ^  5  &C.      - 

LESSON  XXX. 

TRANSFORMATION  OF  FRACTIONS. 

1.  I  have  reserved,  for  a  separate  lesson,  the  third  pro- 
position, corresponding  to  that  in  division,  on  account  of 
its  great  importance  and  frequent  application  in  arith- 
metical operations.  It  forms  the  basis  of  several  systems 
of  calculations,  especially  propositions  and  cancelling, 

PROPOSITION    III. 

If  the  numerator  and  denominator  of  a  fraction  be 
multiplied^  or  divided,  hy  the  same  number^  the  value  of 
the  fraction  itself  will  remain  the  same.  Its  form  alone 
will  be  changed. 


104  LESSON   XXX. 

This  is  the  same  proposition  as  that  in  Lesson  XXIV.,  4, 
by  changing  the  words  dividend,  divisor,  and  quotient, 
into  numerator y  denominator,  diiid  fraction, 

EXAMPLES. 

If  both  terms  of  the  fraction  f  be  multiplied  by  2,  it  becomes  f . 
If  they  be  divided  by  2,  it  becomes  -J* 
And  it  is  evident  that  X  =  1  =z  4^. 

You  get  one  half  of  an  apple,  v^'hether  as  one  half,  two  quar- 
ters, or  four  eighths. 

2.  In  order,  however,  to  explain  this  important  princi- 
ple fully,  independently  of  other  propositions,  let  us  take, 
as  an  example,  the  fraction 

3 

and  multiply  its  two  terms  by  3 ;  we  will  change  it  thus 
into 

A- 

The  first  form  of  the  fraction  expresses  that  the  princi- 
pal unit,  an  apple,  for  instance,  has  been  divided  into 
4  small  units  or  parts,  and  that  three  of  them  are  the 
share  taken. 

But,  in  the  second  form,  each  fourth  itself  is  now  sub- 
divided into  3  parts,  so  that  the  whole  makes  12  parts; 
out  of  which  the  three  pieces,  constituting  the  particular 
share  taken,  would  form  9  parts,  or  -^^  of  the  whole, 
which  is,  therefore,  the  same  thing  as  ^  expressed  in 
units  made  3  times  smaller,  but  whose  reduced  size  is 
compensated  by  their  number;  there  being  3  of  the 
smaller  units,  for  each  one  of  the  larger. 

3.  This  proposition  shows  that  the  value  of  a  frac- 
tion does  not  depend  at  all  upon  the  actual  magnitude  of 
its  terms,  hut  only  upon  their  numeral  relation, 

4.  This  relation,  wbicb  is  the  quotient  of  the  second 
hy  the  first,  is  usually  called  Ratio. 

Thus  the  fractions  |j  |,   A,  -^■^, T%,^W> 

are  all  equivalent,  because  the  denominator  of  each  is  equal  to 


TRANSFORMATION    OF    FRACTIONS.  10^ 

twice  its  numerator,  though  the  magnitudes  of  the  corresponding 
terms  are  so  different.     This  ratio  is  2  for  every  one  of  them. 

Likewise,  f,  y®^'  il>  2^>  ^^''  ^^^  "^^  equivalent  to  |,  be- 
cause they  result  from  the  multiplication  of  its  two  terms  re* 
spectively  by  2,  3,  4,  5,  &c.     The  ratio  here  is  4.* 

5.  This  shows  also  that  the  number  of  forms  under 
which  a  fraction  may  he  expressed^  without  altering  its 
value,  is  as  infinite  as  the  numbers  by  which  both  of  its 
terms  may  be  multiplied, 

A  change  of  form  is  frequently  necessary  to  facilitate 
arithmetical  operations  upon  fractions. 

This  change  or  transformation  may  be  made  in  two 
ways,  either  by  augmentation  of  the  terms  or  by  reduc- 
tion. 

*  Beginners  are  apt  to  think  that  adding  the  same  number  to,  or 
subtracting  it  from,  the  two  terms  of  a  fraction,  does  not  change  its 
value.     This  is  a  mistake. 

Adding  the  same  number  to  both  terms,  increases  the  value  of  the 
fraction. 

Subtracting  lessens  it. 

Take,  for  example,  -~-^ ;  it  is  y^vr  less  than  unity,  which  is  y|- 
Now  add  any  number,  6,  to  both  terms  of  he  given  fraction;  you 
get  ll",  which  is  larger  than  y^'  ^^"^^^  i^  dillers  only  y^^  from  unity, 
which  is  yf  • 

On  the  contrary,  deduct  6  from  both  terms,  and  you  get  -J-' 
which  is  smaller  than  J 

It  will  be  readily  perceived,  in  general,  that  the  difference  be- 
tween any  fraction  and  unity,  is  always  a  supplemental  fraction, 
having  the  same  denominator,  and  whose  numerator  is  the  difference 
between  the  denominator  and  the  numerator  of  the  original  fraction. 
But  now  this  difference  is  not  changed  by  an  equal  addition  to  both 
terms ;  so  that  the  numerator  of  the  supplemental  fraction  is  inva- 
riable, while  its  denominator  is  increased.  It  follows  that  the  sup- 
plemental fraction  is  diminished ;  and,  consequently,  the  given 
fraction,  by  addition  to  both  its  terms,  approaches  nearer  to  unity^ 
and  thereby  is  increased. 

A  similar  mode  of  reasoning  would  show  that  equal  subtraction 
frOm  both  terms  diminishes  a  fraction.  In  the  case  of  an  improper 
fraction,  the  effect  is  reversed  :  addition  decreases,  and  subtraction 
increases  it ;  and  it  would  be  demonstrated  in  the  same  way,  by 
comparison  to  unity. 

Both  cases  may  be  comprised  in  one,  by  saying  that  equal  addition 
to  both  terms  of  any  fraction  mahes  it  approach  nearer  to  unity;  and 
equal  subtraction  makes  it  recede  further  frorn  it.. 


106  LESSON   XXX. 

FIRST  TRANSFORMATION— BY  AUGMENTATION. 

6.  The  object  in  enlarging  the  terms  of  a  fraction,  with- 
out altering  its  value,  is  generally  to  express  it  with  a 
larger  denominator,  or,  in  other  words,  in  smaller  units. 

As  when  -J  is  called  ^ ',  the  eighth  part  of  a  thing  being  only 
one  fourth  part  of  its  half. 

Or,  also,  when  -3-  of  a  foot  is  called  jj  >  that  is,  4  inches^  which 
are  smaller  units  than  the  foot. 

The  enlarged  form  of  a  fraction  is  obtained  by  mulli' 
plying  both  terms  by  such  a  number  as  will  change  the 
first  denominator  into  the  second. 

The  number  we  must  multiply  by,  is  evidently  the 
quotient  of  the  second  denominator  by  the  first. 

If,  for  instance,  we  wish  to  express  the  fraction  -f,  with  the 
denominator,  36,  we  multiply  both  terms  by  12,  which  is  the 
quotient  of  36  by  3. 

The  transformed  fraction  is  then  f -g-  =  f* 

7.  It  follows,  also,  from  this,  that  the  corresponding 
terms  of  two  or  more  equal  fractions  may  be  added  to- 
gether, viz.,  numerators  to  numerators^  and  denominators 
to  denominators,  and  the  result  ivill  still  be  the  same 
fraction  merely  transformed. 

For  example,  take  J  and  -|- ;  it  is  clear  that  =  -|-  still. 

«     f  and  -g^,  by  such  an  addition,  will  make 

^   '^     =  -I- ;  w'hich  is  another  form  for  4» 
3  +  6        ^'  ^ 

This  evidently  amounts  to  multiplying  both  terms  by 
the  same  number;  in  the  first  instance  by  2,  in  the 
second  by  3 ;  or,  in  other  words,  dividing  the  sum  of 
several  equal  dividends  by  the  same  number  of  divisors, 
which  must  give  the  same  quotient  as  one  dividend  by 
one  divisor. 

8.  Sometimes  it  is  useful  to  put  a  whole  number  under 
fractional  form.  This  is  done  by  multiplying  it  by  the 
intended  denominator,  over  which  the  product  is  placed 
as  a  numerator. 


TRANSFORMATION   OF    FRACTIONS.  107 

5  may  thus  be  made  ^  ;  since  15  divided  by  3,  is  5.  This 
amounts  to  changing  the  size  of  the  units ;  5  yards,  for  example, 
into  15  feet, 

SECOND  TRANSFORMATION— BY  REDUCTION. 

9.  The  object  of  reducing  the  form  of  a  fraction  is 
generally  to  simplify  it.  The  reduction  of  a  fraction  to 
simpler  terms,  is  made  by  dividing  both  terms  by  some 
common  factor. 

This  amounts  to  expressing  the  value  of  the  fraction  in 
larger  units. 

As  when  f  is  called  ^  j  the  half  of  a  thing  being  double  one 
quarter. 

Or,  also,  when  6  inches,  otherwise  j2  ^^  ^  f^^^?  ^^^  called  ^ 
a  foot. 

In  the  fraction  |-|,  for  example,  it  is  easily  discovered 
that  2  is  a  common  factor.  We  may,  therefore,  divide 
both  terms  by  it,  and  thus  reduce  the  fraction  to 

30 

a  simpler  form  than  the  preceding.  This  also,  by  a 
second  division  of  both  terms  by  2,  may  be  reduced  to 

and,  finally,  by  striking  out  the  common  factor,  3,  from 
both  numerator  and  denominator,  we  simplify  still  fur- 
ther to 

5 

which  is  not  susceptible  of  reduction,  since  there  is  no 
common  factor  between  5  and  6.  This  is  consequently 
the  simplest  form  of  the  fraction. 

10.  The  change  of  form  of  a  fraction  is  properly  a 
change  of  the  relative  vnits  in  which  it  is  expressed,  in 
reference  to  a  principal  unit  of  comparison. 

When  we  transformed  f  into  f  g^,  we  expressed,  in  thirty -sixth 
parts  of  the  principal  unit,  the  value  af  the  fraction,  which  wes 
at  first  giT«ai  ia  thirds  of  the  same. 


108  LESSON   XXX. 

In  the  example  of  reduction,  the  units  were,  at  first,  each  one 
Sicventy -second  part  of  unity,  and  the  same  value  was  at  last  ex- 
pressed in  sixths  ;  |.  being  equal  to  5.0. 

11.  Sometimes  it  may  be  desired  to  transform  a  frac- 
tion, in  reference  to  a  given  numerator.  In  that  case, 
the  rule  is  similar  to  that  in  reference  to  the  denomi- 
nator. 

Multiply^  or  divide,  hotli  terms  hy  the  numher  which 
would  change  the  first  numerator  into  the  second  (that  is, 
l)y  their  quotient). 

For  example,  to  change  -f^-  into  a  fraction  having  the  numera- 
tor 72,  multiply  both  terms  by  3  :  the  transformed  fraction  will  be 

To  reduce  it  to  one  having  the  numerator  4,  divide  both  terms 
by  6,  because  4  X  6  =  24.     The  result  will  be 

4. 

'6  7 

and  it  is  evident  that  i'i  f  f  5  y^j  are  all  equivalent. 

Questions. — If  the  two  terms  of  a  fraction  are  multiplied  by 
the  same  number,  what  effect  has  it  on  its  value  ?  What,  if 
divided  by  the  same  number  ?  Why  ?  Give  examples.  Repeat 
the  proposition.  Does  the  value  of  a  fraction  depend  on  the  mag- 
nitude of  its  terms  ?  What  does  it  depend  on  ?  What  is  meant 
by  ratio  ?  Can  a  fraction  be  expressed  in  different  ways  ?  How 
niany?  In  how  many  ways  may  a  fraction  be  transformed? 
What  are  they?  What  does  transformation  by  augmentation 
amount  to  ?  How  do  you  give  to  a  fraction  a  larger  denominator 
without  changing  its  value?  Give  examples.  How  do  you 
transform  a  fraction  by  reduction  ?  Give  examples.  When  is  a 
fraction  reduced  to  its  lowest  terms  ?  To  its  simplest  expression  ? 
How  would  you  increase  or  diminish  the  numerator,  without 
altering  the  value  of  the  fraction  ?  What  is  the  ratio  of  9  to  18  ? 
A71S.  2,  Of  16  to  48?  Of  46  to  16?  Of  11  to  15?  Of  15  to 
11?     Of  25  to  5?     Of  5  to  25?  &c. 

EXERCISES. 

1.  Transform  the  fraction  f  into  fractions  having  for  denomi- 
nators 15;  20;  35;  65-,  60;  75;  85;  105;  250;  155;  2550,  &c. 

2.  Transform  the  fraction  y  so  that  its  denominator  may  be 
successively  4,  5,  9,  15,  20,  24  times  larger. 


PERMUTATION   OF    FACTORS   AND    DIVISORS.  109 

3.  Change  7  into  a  fractional  expression,  with  the  denominator 
4,  7,  11,  25. 

4.  Express  8  in  units,  5,  6,  8,  10,  12  times  smaller. 

5.  Express  the  fraction  f  in  units,  2  j  Sj   5;    6;   10;   11;   12 
times  smaller. 

6.  Reduce  the  fractions  #;  }§;  Z^;   }-|;  ||;  ^^  5  If  5  ||j 
9  ;   96;  -L^j^;  Ji-,  to  their  simplest  expressions. 

7.  Express   the  fractions  y^'  H'  ii"'  ■J"!'  i^^  ""i^s  3  times 
larger. 

8.  Express  the  fractions  j^i  ^^i  |§;  IJ;  -fl^  in  units  5 times 
larger. 

9.  Express   the  fraction   yj  with  the  numerators  12,  21,  39, 
54. 

10.  Express  the  fraction  |-^  with  the  numerators   15,  10,  5, 
3,1, 


CHAPTER  V. 

CONTAINING    RULES    RELATIVE    TO    THE    FACTORS    AND 
DIVISORS    OF    NUMBERS,    AND    CANCELLING. 

N.  B. — The  whole  of  this  chapter  may  be  omitted  with  begin- 
ners. 

LESSON  XXXI. 

PERMUTATION  OF  FACTORS  AND  DIVISORS. 

The  answer  to  many  questions  in  arithmetic,  requires 
several  simultaneous  multiplications  and  divisions.  It  is, 
therefore,  important  to  establish  clearly  the  following 
propositions,  in  regard  to  the  order  in  which  they  may 
be  performed,  before  we  introduce  combined  operations. 

PROPOSITION   I. 

1.   When  several  numbers  are  to  be  multiplied  toge* 
tker,  the  product  will  be  the  same,  in  whatever  order  the 
multiplications  are  made. 
10 


110  LESSON   XXXI. 

This  principle  has  already  been  demonstrated  for  two 
factors ;  it  may  be  illustrated  for  three,  by  the  following 
example. 

Question, — There  are  5  rows,  of  4  baskets  each,  and  in  each 
basket  6  peaches  :  how  many  peaches  are  there  in  all  ? 

We  may  arrive  at  the  result  in  several  ways. 

1st.  We  may  get  at  the  number  of  baskets  by  multiplying  the 
4  baskets  in  one  row  by  the  number,  5,  of  rows, 
which  is       ...4X5]  and  then  multi-  ]  4  X  5  X  6  peaches, 
or  the  number  of  rows,  I  ply  the  product  ! 

5,  by  the  4  baskets  in  [  by  the  6  peaches  [ 

each 5  X  4  J  in  each  basket.    J  5  X  4  X  6 

2d.  We  may  count  the  peaches  in  a  row  of  baskets, 

equal  to  .     .     .     .     4X61  peaches,  and  mul-]  4  X  6  X  5i 
I  tiply  this  number  I 
•  I  by  the  number  off 

or 6  X  4  J  rows.  J  6  X  4  X  5 

3d.  We  may  count  the  peaches  in  the  baskets,  at  the  head  of 
each  of  the  5  rows, 

equal  to  .     .     .     .     5  X  6  i  and  then  repeat  it  i  5  X  6  X  4 


!  }  and  then  repeat  it  i  5  ; 

I  by  the  number,  4,  I 
>  \  of  such  rows.  \  6  ; 


or 6  X  5  3  of  such  rows.  ^6x^X4 

All  of  which  must,  of  course, .  give  the  same  number  of 
peaches,  120. 

2.  A  popular  illustration  like  this,  however,  is  not 
always  practicable ;  and  it  is  proper  to  demonstrate  every 
principle  by  abstract  and  more  mathematical  methods. 
Here  we  may  consider  that,  according  to  what  has  been 
previously  demonstrated  (XII.,  5),  the  same  result  is  finally 
obtained  in  a  multiplication,  whether  we  multiply  the 
multiplicand,  the  multiplier,  or  the  product,  by  a  given 
number. 

Therefore,  having  to  multiply  the  product  4  X  ^  by  6,  we  may 
make,  first, 

the  product  4  X  5  (  and  then  multiply  it  J  by  6,  to  get  (  4  X  5  X  6 

or  its  equal,        5X4^  «  \      "      or)5X4X6 

Otherwise,  we  may  first 
multiply  the  mul-         "j  f  by  5,  to  get  "j  4  X  6  x  5 

whSthe  samt  ^^  ^  f  ^^  *^^^  "^^"P^^  ]  \ 

as  .    .    .    .    6by4j  «  (^     «     or    J  6  X  4  X  5 


PERMUTATION   OP   FACTORS   AND   DIYISORS.  Ill 


Or,  finally,  we  may  begin  with  multiplying 
the  multiplier    5  by  6  i  and  then  multiply     (  by  4,to  get  C  5  X  6  X  4 
which  is  the  same  >  <  < 

as  ....     6  by  5  i  '"  ^      "     or     ^  6  X  5  X  4 

Showing  every  possible  combination  of  three  factors 
for  the  same  product. 

3.  The  principle  being  now  established  for  three,  may 
be  easily  extended  to  any  other  number  of  factors. 
Let  us  take,  as  an  example,  the  product  of 

2x4x5x7x8x65 

in  which  6  multiplies  the  product  of  all  the  preceding 
factors. 

If,  now,  we  transfer  6  to  any  other  place ;  the  third, 
for  example,  we  shall  have 

2x4x6x5x7x8. 

Let  us  make  the  products  which  precede  and  follow  6. 
We  shall  have,  then,  three  numbers}  and,  according  to 
what  has  just  been  explained, 

8x6x280  =  8x280x6; 
or,  by  restoring  the  factors, 

2x4x6x5x7x8=2x4x5x7x8x6; 

and,  since  the  same  mode  of  reasoning  would  apply  to 
any  other  place  chosen  for  6,  and  also  to  any  other  fac- 
tor, we  might  form,  in  that  way,  all  the  possible  changes 
the  number  of  factors  admits  of,  without  altering  the 
product.     Therefore, 

The  permutation  of  the  factors  does  not  alter  the 
product. 

And  we  may,  in  practice,  adopt  the  order  which  will 
best  facilitate  the  operations. 

For  instance,  if  we  had  to  multiply 

3x125x5x7x8x2, 

the  order  in  which  the  numbers  are  presented  would  be 
unfavorable ;  but  the  operation  might  be  greatly  simplified 


112 


LESSON   XXXI. 


by  a  change  of  this  order :  for,  it  would  be  readily  per- 
ceived that  2x5  =  10;  8x125  =  1000;  3x7  =  21, 
and  that  the  entire  product  is  210,000.  In  this  way,  the 
opeiation  could  be  easily  perfornned  mentally. 

CONTRACTION  IN  MULTIPLICATION. 

4.  It  follows  from  this  proposition,  that,  instead  of 
multiplying  hy  the  wliole  multiplier^  xoe  may  multiply 
successively  hy  its  component  factors;  since  the  multi- 
plicand and  all  the  factors  of  the  multiplier  must  form 
the  product  in  whatever  order  we  may  choose  to  take 
them. 

For  exannple,  if  we  had  to  multiply    365  by  96, 
we  might  multiply,  first,            -        -     365 
by       -        - 8_ 

=    2920 
and  then  this  product  by  -        -       12 

=    35,040 

This  is  called  contraction  in  multiplication.  It  may 
facilitate  mental  operations. 

Questioiis. — What  is  the  effect  of  permutations  of  the  factors 
upon  the  product  ?  Demonstrate  it  for  three  factors.  For  any 
number.  Repeat  the  first  proposition.  What  is  contraction  in 
multiplication  ?     Give  an  example. 

EXERCISES   ON    MULTIPLICATION    BY   CONTRACTION. 

1.  43,102  X  66=  2,844,732.  6.  24,683  X375=  9,256,125. 

2.  871,075  X  42  =  36,585,150.         7.  2,947  X  336=  141,456. 

3.  526,473  X  144=  75,812,112.  8.  3,726  X  132  =  491,832. 

4.  12,071  X  99=  1,195,029.  9.  3,748  X  147  =  550,956. 

5.  47,701  X  242=  11,543,642.  10.  2,164  X  153=  231,092. 

LESSON  XXXII. 

PROPOSITION    II. 

1.   When  the  divisor  is  composed  of  several  factors. 


PERMUTATION    OF    FACTORS    AND    DIVISORS.  113 

the  quotient  will  he  the  same^  whether  we  divide  hy  the 
whole  divisor  at  once,  or  successively  by  its  component 
factors,  in  any  order  whatever. 

For,  the  quotient  is  another  factor  which,  multiplied 
by  those  of  the  divisor,  must  produce  the  dividend.  If, 
then,  we  divide  by  any  one  of  the  factors,  the  result 
contains  yet  the  quotient  and  the  other  factors ;  and  a 
succession  of  divisions,  by  the  various  factors,  will  reduce 
the  result  at  last  to  the  quotient  alone. 

Let  it  be  proposed,  for  instance,  to  divide  168  by  24. 
The  quotient  would  be  7,  and  24  being  6X4,  we  bave 

16S  =  6X4X7. 

Consequently,  if  we  divide  168  first  by  4,  tbis  factor  is 
removed  from  tbe  dividend ;  then  6  and  7  remain,  and  the 
result  of  the  first  division  is 

42  =  6X7. 

Another  division,  by  6,  will  now  reduce  tbe  first  result 
to  tbe  required  quotient,  7. 

This,  in  fact,  amounts  to  dividing  the  dividend  and 
divisor  by  the  same  number,  which  we  know  does  not 
alter  the  quotient  (XXIV.,  4). 

2.  Divisions  are  sometimes  performed  in  this  way,  by 
the  successive  component  factors  of  the  divisor. 

For  instance,  1728  might  be  divided  by  72,  by  means 
of  two  divisions  j  the  first  by  8,  the  second  by  9. 


OPERATION. 


Dividend       .     1728 


1st  quotient   •       216 


1st  divisor. 
2d  divisor. 


last  quotient  .         24 

This  method  is  called  Contraction  in  Division, 

Contraction  is  more  useful  in  division  than  in  multi- 
plication, because  it  substitutes  short  for  long  division. 

3.  It  is  evident  that  the  order  of  the  divisions  does  not 
10*  H 


114'  LESSON  xxxir. 

affect  the  quotient :  since  permutation  of  its  factors  will 
not  change  the  amount  of  the  divisor. 

4.  From  what  precedes,  we  may  also  conclude  that, 
whatever  may  he  the  place  of  a  factor^  in  a  product  to 
he  formed^  we  divide  the  product  hy  this  factor  hy  striJcing 
it  out. 

Thus,  if  we  want  to  divide  by  6?  the  indicated  pro- 
duct, 

4x6x5x7, 

the  quotient  will  be  4  X  5  X  7  j  since  this  multiplied 
by  6,  would  reproduce  the  same  amount. 

5.  Observe  that,  in  striking  out  a  factor,  you  divide  by 
it;  and  that,  since  a  number  divided  by  itself  gives  1,  it 
is  understood  that  1  remains  in  lieu  of  the  factor  struck 
out.  But  this  figure,  as  a  factor,  is  generally  omitted, 
because  it  has  no  effect  on  the  product,  and 

4x5x7     is  the 
same  as      ...    4x5x7      X 1. 

PROPOSITION    III. 

6.  Wlien^  in  an  operation^  a  numher  is  to  he  multiplied 
and  also  divided,  it  is  immaterial  whether  you  multiply 
or  divide  first. 

For,  the  numbers  to  be  multiplied  form  a  product; 
and  it  amounts  to  the  same  thing,  whether  you  divide 
either  of  the  factors  or  the  product  (XXIII.,  3). 

EXAMPLE. 

Suppose  that  we  have  to  multiply  144  by  17,  and 
divide  by  24.     The  operation  will  be  indicated  thus : 

144X17_.,^^ 
24 

We  might,  by  multiplication,  get  first  2,448,  and  then 
divide  by  24 :  but,  we  may  also  divide  by  24,  and  then 
multiply  the  quotient  6  by  17. 


OF    CANCELLING.  Il5 

Generally,  to  begin  by  division  is  preferable,  because 
it  simplifies  the  work ;  and  it  is  absurd  to  enlarge  a  num- 
ber first,  to  reduce  it  afterwards. 

PROPOSITION    IV. 

7.  When  the  'product  of  several  factors  is  to  he  du 
Tided  by  a  number  composed  also  of  factors,  instead  of 
dividing  the  product  itself,  we  may  divide  successively 
any  one  of  its  factors  by  any  one  of  those  of  the  divisor, 
until  the  division  is  made  as  far  as  practicable. 

Because  we  divide  a  product  by  one  of  its  factors, 
wben  we  strike  out  tbis  factor  (XXXII.,  1  and  4). 

GENERAL    PROPOSITION. 

8.  The  four  preceding  propositions  may  be  condensed 
into  the  following  single  one : 

When  several  multiplications  and  divisions  are  com* 
hined  in  the  same  operation,  it  is  immaterial  in  what 
order  you  multiply  and  divide, 

OF  CANCELLING. 

9.  The  subject  of  this  lesson  is  very  important.  It  is 
one  of  the  principles  on  which  are  based  the  simplifica- 
tions called  cancellings ;  which  consist  in  striking  out 
from  a  dividend  and  divisor,  or,  in  other  words,  numera* 
tor  and  denominator,  all  common  factors. 

For  cancelling,  we  must  be  able  to  discover  the  sim- 
pler factors  of  some  of  the  numbers.  There  are,  for  this 
purpose,  several  rules  on  the  divisibility  of  numbers, 
which  it  is  important  to  be  thoroughly  conversant  with, 
and  which  form  the  subject  of  the  thirty-fourth  lesson. 

In  cancelhng,  if  you  strike  a  factor  out  of  a  number,  cross  the 
number,  and  write  above  or  below  it  the  quotient  which  remains 
in  lieu  of  the  number. 


116  LESSON    XXXII. 


72  X  45  X  16    I      Cancelling  4  out  of  20  and  72;  5  out  of  20 
20  X  12  X  24    I  and  45 ;  and  8  out  of  16  and  24 :  the  arrange- 
ment will  be     18       9         2 


Now,  we  may  still  cancel  6 
out  of  18  and  12;  then  the  re- 
maining 2  of  12,  with  2  of  the 
numerator,  and  also  3  with  9. 


^0Xl2X^^ 
3 
This  second  change  will  give      3        3 

With  a  little  practice,  the  whole  cancelling  might  have  been 
done,  and  9  obtained  at  once. 

Qiiestions. — Repeat  the  second  proposition;  the  third;  the 
fourth.  Will  the  quotient  be  the  same,  if  you  divide  successively 
by  the  component  factors  of  the  divisor  ?  Why  ?  What  is  con- 
traction in  division?  Give  an  example.  Will  the  order  of  the 
factors,  in  contracted  division,  affect  the  quotient  ?  Why  ?  What 
effect  is  produced  on  the  indicated  product  of  several  factors,  if 
one  is  struck  out  ?  What  is  left  in  its  place  ?  Is  it  necessary  to 
write  down  the  quotient  1  as  a  factor  ?  If,  in  the  same  question, 
several  multiplications  and  divisions  are  to  be  made,  what  will 
be  the  effect  of  changing  the  order  of  these  operations  ?  May  we 
divide  any  factor  of  the  dividend  by  any  factor  of  the  divisor  ? 
Why  ?  Repeat  the  general  proposition  for  all  the  preceding. 
What  is  cancelling  ?  Where  do  you  set  down  the  quotient  of  a 
number,  out  of  which  one  or  more  factors  have  been  cancelled  ? 
If  a  whole  number  is  cancelled,  what  do  you  set  down  in  its 
place  ?     What  is  understood  to  occupy  its  place  ? 

EXERCISES. 

1.  Prove  the  multiplications  of  the  preceding  lesson  by  con- 

traction in  division. 

Divide  by  contraction — 

2.  1,141,365,312  by  36  =  31,704,592. 

3.  2,142,273,888  by  72  =  29,753,804. 

4.  684,323,328  by  96  =  7,128,368. 

5.  17,326,526,400  by  108  =  160,430,800. 

6.  81,623,150,400  by  1320  =  61,835,720. 

7.  999,999,999  by  81  =  12,345,679. 


ADDITIONAL   DEFINITIONS.  117 

Perform  the  following  multiplications  and  divisions  by  cancel- 
ling: 

20  XJ5^X8  __  ^ 
30  X  10 

_  1  2X3X5X7X8  _  ^q 

2*     9  V  .q  V  4  V  /^  ~  5*  ^'     3X4X7X9X11""^^* 


!•         O   vv   O    vy    ^    4*  4. 


2X3     1 

2X3X4   4- 

2X3X4 

2X3X4X5  "~ 

2X3X10_  4 

3*  O  vy   /^  w   1  1     11*  6. 


7X11X27X5         33 


3X5X11""^'*  18X10X14X2        i'^' 

2X4X7X3        _    ., 
^'     2X3X5X8X4X7""  4D' 

6X1X4X2X7X18  _    ^^ 
^'      7X2X3X11X9X25       aTs' 

12X11X10X9X8  _  r.Q2 
^'  2X3X4X5         —       ^* 

45X25X40X10X7X9  _  -^^43 
^^'        20X15X8X6X11  ^4* 


12. 
13. 


500  X  12  X  2900  X  12  X  5  X  3  X  57  ___  5405 1_ 
860X10X1800X7X3  "~     *^3  0i' 

216  X  192  X  11  X7X975X5QX  10X20  __  ^^^  o-j 
140  X  9  X  11  X  546  X  45  X  15  X  25        ~" 


LESSON  XXXIII. 

Before  presenting  the  rules  on  the  divisibility  of  num- 
bers, it  is  necessary  to  introduce  some  additional  defini- 
tions and  remarks. 

ADDITIONAL   DEFINITIONS. 

1.  An  even  number  is  one  that  can  be  divided  exactly 
by  2. 

All  numbers  ending  with  0,  2,  4,  6,  8,  are  even  numbers. 

2.  An  odd  number  is  one  that  cannot  be  divided  by  2. 
All  numbers  ending  with  1,  3,  5,  7,  9,  are  odd  numbers. 

3.  A  prime  number  is  one  which  cannot  be  divided 


118  LESSON   XXXIII. 

without  a  remainder,  by  any  other  number  than  itself  or 
unity. 

1,  2,  3,  5,  7,  11,  13,  17,  19,  23,  29,  &c.,  are  prime  numbers. 

About  one  fourth  of  the  odd  numbers  are  prime.    They 
are  all  one  unit  above  or  one  unit  below  a  multiple  of  6. 


IX  12=  72    one  less  is  71 
one  more 


71  } 

„o  (  both  prime  numbers. 


6  X  11  =  66      66  + 1  =  67,  a  prime  number. 
66— 1=  65  is  not  one. 

There  are  but  few  products  of  6,  which  are  not  within 
one  unit  of  at  least  one  prime  number.* 

'    6  X  20=  120,  is  one  of  the  (120  + 1  =  121,  which  is  11  X  H. 
exceptions.  J  120  —  1  =  119,        "         7  X  17. 

4.  Two  numbers  are  said  to  be  prime  to  each  other^ 
when  no  common  factor  will  divide  both. 

25  and  49  are  prime  to  each  other. 
25  and  45  are  not,  because  5  divides  both. 
Nor  are  21  and  49,  which  have  the  common  factor  7. 

5.  A  factor  common  to  two  numbers,  is  also  called 
their  common  divisor, 

4  is  a  common  divisor  of  48  and  72. 

The  greatest  number  which  will  divide  two  or  more 
numbers,  is  their  greatest  common  divisor. 

In  the  preceding  example,  4  is  a  common  divisor ;  but  not  the 
greatest  between  48  and  72,  which  is  24.  The  quotients  being 
respectively  2  and  3,  which  have  no  factors,  24  is  evidently  the 
greatest  between  48  and  72. 

In  the  preceding  lesson,  instead  of  reducing  -f-f  to  its  lowest 
terms,  |-,  by  successive  operations,  it  might  have  been  done 
quicker  by  dividing  both  terms  by  their  greatest  common  divisor^ 
12. 

It  is  generally  easy  to  reduce  a  fraction  by  successive 
divisions;  but  sometimes  it  requires  the  finding  of  the 

*  Algebraically  expressed,  &n  i  1,  it  contains  all  the  prime 
numbers. 


ADDITIONAL   DEFINITIONS.  119 

greatest  common  divisor,  for  which  a  method  will  be 
given  hereafter  (Lesson  XXXVIII.). 

6.  We  know  that  a  fraction  is  reduced  to  its  simplest 
expression,  or  lowest  terms,  ivhen  its  two  terms  are  prime 
to  each  other, 

7.  A  fraction  thus  reduced,  and  which  cannot  be  sim- 
plified, is  called  an  irreducible  fraction. 

8.  A  multiple,  it  has  been  already  said,  is  a  number 
which  contains  another  as  a  factor. 

The  factor  is  sometimes  also  called  an  aliquot  part  of 
its  multiple. 

12  is  a  multiple  of  2,  of  3,  of  4,  and  also  of  6,  because  each  of 
them  will  divide  it  exactly. 

2,  3,  4j  6,  are  aliquot  parts  ;  that  is,  factors  of  12. 

It  will  be  remarked  that  the  same  number  may  have 
an  infinite  number  of  multiples :  6,  24,  90,  600,  Q,QQQy 
&c.,  are  all  multiples  of  6. 

But  it  has  a  limited  number  of  aliquot  parts :  12,  for 
instance,  has  only  the  aliquot  parts  2,  3,  4,  6. 

9.  The  same  number  may  be  the  multiple  of  many 
others :  12  has  already  served  as  an  example. 

A  number  which  is  the  multiple  of  several  numbers  is 
called  their  common  multiple. 

Thus,  12  is  the  common  multiple  of  2,  3,  4,  and  6. 
60,  120,  240,  360,  are  all  common  multiples  of  2,  3,  4,  6,  12, 
15,  &c. 

Among  the  common  multiples  of  several  numbers, 
there  is  evidently  one  smaller  than  all  the  others,  and 
which  is  just  large  enough  to  contain  them.  This  is 
called  the  least  common  multiple. 

In  the  last  example,  60  is  the  smallest  common  multiple ;  for 
no  smaller  number  can  be  found  to  contain  4  and  15;  and,  con- 
taining these,  it  of  course  contains  their  factors,  2  and  3. 

10.  The  number  whichi  expresses  the  relation  of  a  mul- 
tiple to  its  factor,  is  also  called  their  Ratio  (XXX.,  4). 

11.  The  term  power  is  frequently  used  to  express  the 


120 


LESSON   XXXIV. 


number  of  times  that  the  same  factor  appears  in  a  pro- 
duct.    Thus  : 

2X2  is  the  second  -power  of  2.  It  is  generally  written  2-. 

3x3  is  the  s^zon^  power  of  3.  "  "         S^. 

3  X  3  X  3  X  3  is  the  fourth  poioer  of  3.  «  "         3^ 

7X7X7XVX7  is  the  fifth  j9o«;er   of  7.         «  "        75. 

Questions. — What  is  an  even  number ?  An  odd  number?  A 
prime  number  ?  Are  there  many  prime  numbers  ?  If  you  multi- 
ply a  number  by  6,  and  add  or  subtract  1,  will  it  often  make  a 
prime  number  ?  When  are  numbers  prime  to  each  other  ?  Which 
are  the  prime  numbers  in  43,  45,  51,  59,  61,  63,  67,  73,  77,  81, 
89,  91,  93,  99  ?    Are  5  and  7  prime  to  each  other?    Are  6  and  9  ? 

9  and  11?  12  and  15/  6  and  25?  23  and  32?  45  and  49? 
What  is  a  common  divisor?  The  greatest  common  divisor? 
When  is  a  fraction  reduced  to  its  simplest  expression  ?  What 
is  it  then  called?  What  is  a  multiple?  Of  what  numbers  are 
24;  36;  63;  96,  &c.,  multiples?  What  is  the  least  common 
multiple  of  2,  3,  4  ?  Of  2,  5,  6  ?  Of  4,  8,  6,  3  ?  Of  5,  3,  6,  2  ? 
&c.?  What  is  an  aKquot  part?  What  are  the  aliquot  parts  of 
12,  36,  48  ?  What  term  may  express  the  relation  between  mul- 
tiple and  factor  ?     What  number  expresses  the  rekition  between 

10  and  2?  12  and  3?  18  and  9?  28  and  7?  66  and  11?  90 
and  10  ?  &c.  What  is  the  ratio  of  4  and  8;  5  and  20  j  6  and  42  ? 
&c.     What  is  a  power  ? 


LESSON  XXXIV. 
DIVISIBILITY  OF  NUMBERS. 

To  simplify  and  facilitate,  in  many  cases,  multiplica- 
tion, division,  the  reduction  and  combination  of  fractions  5 
to  detect  errors  in  those  operations ;  and,  in  general,  to 
acquire  readiness  and  dexterity  in  numbers,  it  is  necessary 
that  the  arithmetician  should  be  able  to  discover  with 
ease  the  simple  factors  of  numbers.  I  will  present  here, 
therefore,  some  of  the  most  useful  rules  for  this  purpose. 

1.  2  divides  every  even  number, 

2.  3  divides  exactly  any  number  when  the  sum  of  its 
figures  is  divisible  by  3. 

Examples. — 42  ;  72;  252  are  divisible  by  3,  because  4+2  =  6; 
7  +  2  =  9;  2  +  5  +  2  =  9;  and  6  and  9  are  multiples  of  3. 


DIVISIBILITY    OF    NUMBERS.  121 

3.  4  divides  any  number  the  last  two  figures  of  which 
make  a  multiple  of  4. 

624;  476;  1548,  are  all  divisible  by  4;  because  24;  76;  48, 
are  thus  divisible :  146  is  not,  because  46  is  not  a  multiple  of  4. 

4.  5  divides  any  number  which  ends  with  0  or  5. 
20;  35;  60;  175;  &C.5  are  all  divisible  by  5. 

5.  6  divides  any  even  number  divisible  by  3  (2). 

36;  42;  96;  12,456,  &c.,  are  divisible  by  2  and  3;  and,  conse- 
quently, by  6. 

6.  7  divides  any  number  when  twice  the  figure  of 
units.,  subtracted  from  the  other  fart  of  the  number, 
leaves  for  a  remainder  0,  or  a  multiple  of  7. 

Thus :    84  is  divisible  by  7,  because  2X4  =  8. 

91  '*  7,  because  tw^ice  1  subtracted  from  9, 

leaves  7. 
343  «  7,  because  2  X  3  =  6,  and  34  —  6 = 28, 

a  multiple  of  7. 
In  2,401,  we  might  subtract  twice  1  from  240,  and  then  try  also 
if  the  remainder,  238,  is  divisible  by  7,  by  saying  2X8=16, 
and  23 — 16  =  7;  which  would  prove,  at  last,  the  divisibility  of 
2,401. 

But  this  tedious  process  is  not  necessary  with  large 
numbers;  and  we  may  bring  down  the  test  to  three 
figures,  by  using  the  following  rule  : 

When  the  number  exceeds  three  figures,  divide  it  into 
periods  of  three  figures,  and  subtract  the  sums  of  the 
alternate  periods  from  each  other  ;  then,  if  the  remainder 
is  0,  or  a  midtiple  ofl,  the  number  is  divisible  by  7. 

If  after  the  test  has  been  reduced  to  three  figures^  the 
divisibility  is  not  apparent,  use  the  first  rule. 

EXAMPLES. 

The  number  801,801  is  divisible  by  7,  because  the  two  periods 
are  equal. 

60,315,325 

also  is  divisible,  because  60  +  325  —  315  =  70;  which  is  an  evi- 
dent multiple  of  7. 
11 


122  LESSON  xxxiy. 

Again,  try     .     .    "39,655 
by  the  second  rule,   655 — 39  =  616,  whose  divisibility  is  not 
evident. 

Then  by  the  first  rule,  2  X  6  =  12,  and  61— 12  =  49,  a  multi- 
ple of  7.     Hence,  39,655  is  divisible  by  7. 

7.  8  divides  any  number  the  three  last  figures  of 
which  form  a  multiple  ofS, 

1,568  is  divisible  by  8,  because  568  is  evidently  divisible,  since 
its  parts  560  and  8  are  separately  so. 

8.  9  divides  any  number  the  sum  of  whose  figures  is 
divisible  by  9. 

81 ;  243 ;  747 ;  28,548,  are  all  divisible  by  9,  because  the  sums 
of  their  figures  are  respectively  9  j  9 ;  18 ;  27 :  which  are  mul- 
tiples of  9. 

The  reason  of  this  rule  is  so  simple  that  it  may  be  ex- 
plained here.  It  is  well  known  that  the  remainder  of 
the  division  of  10  j  100  j  1000,  &c.,  by  9,  is  1:  conse- 
quently, the  remainder 

Of  20;  200;  2000,  &c.,  .         .         is     2. 

Of  30;  300;  3000,  &c.,  .         .         «      3, 

&c.  &c. 

Or,  in  other  words,  the  remainder  of  the  division  of  a 
number  of  any  order  by  9,  is  its  significant  figure. 

Thus,  in  28,548,  which  is  20,000  +  SOOO  +  500  +  40  +  8, 
the  remainders  of  the  division  i 

of  each  order  by  9  are  sepa-  S-  2,8,5,4,8, 
rately  3      .  .  . 

consequently  the  aggregate  remainder  is  their  sum  27,  which 
being  divisible  by  9,  it  follows  that  the  given  number  itself  is  so 
likewise. 

When  the  aggregate  remainder  is  not  divisible  by  9,  its 
own  remainder  is  also  that  of  the  given  number.  This 
second  remainder  is  generally  the  sum  of  the  figures  of 
the  aggregate  remainder. 

Thus,  in  87,865,  the  aggregate  remainder  is  8+7+8  +  6  +  5 
s=  34 ;  which,  divided  by  9,  gives  a  final  remainder,  3  +  4  =  7, 
which  is  also  that  of  the  division  of  87,865  by  9. 

All  these  remarks  apply  also  to  3.  We  will  make  use 
of  this  property  of  9  and  3,  in  the  Appendix. 


DIVISIBILITY   OF   NUMBERS.  123 

9.  10  divides  any  number  which  ends  with  0. 
100         "  "  "        with  two  Os. 

1000         «'  "  "        with  three,  k.c. 

10.  25  and  50  divide  any  number,  the  last  two 
figures  of  which  form  a  number  divisible  by  them. 

275 ;  34,275 ;  9,350,  are  all  divisible  by  25. 
250 ;  550 ;  34,750 ;  9,350,  are  divisible  by  50. 

11.  11  divides  a  number  when  the  difference  between 
the  sums  of  its  alternate  figures  is  either  0  or  a  multiple 

0/11. 

405,812  and  290,718,076  are  both  divisible  by  11 ;  because,  in 
thefirst,  4  +  5  +  1=84- 2;  and,  in  the  second,  2  +  0  +  1+0  +  6 
=  9;  9_^7-|-8  +  7  =  31;  and  31  —  9  =  22;  which  is  divisible 
by  11. 

12.  12  divides  any  number  divisible  by  both  3  and  4), 

288  is  divisible  by  12 ;  because  the  sum  of  its  figures  is  18, 
divisible  by  3  (2),  and  the  two  last  make  a  number  divisible  by 
4  (3). 

Questions. — How  do  you  know  when  a  number  is  divisible  by 
2?  3?  4?  5?  6?  7?  8?  9?  10?  100?  1000?  &c.  25?  50?  11? 
12  ?     Give  a  demonstration  for  9  and  3. 

EXERCISES. 

1.  Which  ofthe  numbers  352;  456;  315;  1,247;  2,677;  5,544; 
7,921,851;  4,817,628,  are  divisible  by  3  and  by  9? 

2.  Which  of  622;  728;  6,982;  4,176;  918,578;  782,496;  34,926, 
are  divisible  by  4?  and  by  8? 

3.  Which  of  245;  373;  504;  924;  2,897;  9,254,154;  3,457,923; 
834,461,  are  divisible  by  7  ? 

4.  Which  of  465;  270;  975;  3,275;  4,050;  3,000;  925;  7,600, 
are  divisible  by  5?  by  10?  by  25?  by  50?  by  100?  by  1000? 

5.  Which  of  649;  7,285;  315,788;  3,259;  7,821,472;  61,809,132; 
9,145,389,  are  divisible  by  11  ? 

6.  Which  of  168;  296;  8,034;  1,236;  2,358,  are  divisible  by 
12?  by  24? 


124  LESSON   XXXV. 


LESSON  XXXV. 


1 .  13  divides  a  numher  when  the  sums  of  its  alternate 
pei'iods  of  three  figures^  subtracted  from  each  other^ 
leave  for  a  remainder  0  or  a  multiple  o/*  13. 

IjOOl  is  divisible  by  13,  because 

1  (of  the  period  of  thousands)  —  1  (of  the  period  of  units)  =  0, 
236,210         <«         «         because236  — 210  =  26,  a  multiple  of  13. 
657,177,209  "        «        because  657+209—177  =  689,  which  is 
divisible  by  13. 

In  a  case  like  the  last,  when  the  subtraction  of  the 
alternate  periods  has  reduced  the  test  to  a  number  of  three 
figures  only,  it  may  not  always  be  readily  perceived 
whether  the  remainder  is  divisible  by  13.  We  must  then 
do  as  was  done  for  7^  use  a  separate  rule  for  three 
figures,  as  follows : 

Multiply  the  figure  of  units  by  9,  or,  ichen  you  can, 
take  one  third  of  it,  and  subtract  bettveen  the  result  and 
the  other  part  of  the  number  ;  the  remainder  must  either 
he  0  or  a  multiple  of  13. 

In  the  remainder,  689,  of  the  preceding  example,  9  X  9  =  81, 
and  81  —  68  =  13 ;  or  else,  |  =  3,  and  68  —  3  =  65  =  5  X  13. 
Again :  104  is  divisible  by  13,  because  4  X  9  — 10  =  26. 

Again :  when  the  lost  two  figures  form  a  number  divi- 
sible by  4,  it  is  sometimes  shorter  to  divide  this  part  by  4f, 
and  subtract  between  the  quotient  and  the  hundreds. 

In  104,  divide    4  by  4,  and  subtract ;  you  get  1  —  1  =    0. 
In  936,  divide  36  by  4,  and  you  get  .     .     .      9  —  9 -=    0. 
In  884,  divide  84  by  4,  v^^hich  gives  21,  and    21  —  8  =  13. 
Showing  that  all  these  numbers  are  divisible  by  13. 

2.  As  regards  17,  cut  off  ttoo  or  three  figures  to  the 
right  of  the  number,  and  then  divide  the  part  thus  cut 
off  by  the  number  of  its  figures  ;  and,  if  the  subtraction 
between  the  quotient  and  the  other  part  of  the  given  num- 
ber leaves  0,  or  a  multiple  of  17,  the  giveji  number  is 
divisible  by  17. 


DIVISIBILITY    OF   NUMBERS.  125- 

In  297,636,  for  example,  cut  off  636,  one  third  of  which  is  212; 
then  subtract  297  —  212  =  85,  and  you  see  that  the  whole  num- 
ber is  divisible  by  17. 

N.  B. — If  the  three  last  figures  were  not  divisible  by  3,  you 
could  always  make  them  so,  by  adding  or  subtracting  17 ;  since 
the  remainder  of  a  division  by  3,  is  always  2  or  1. 

In  1,666,  cut  off  66  ;  of  which,  take  the  half,  33  ;  then  33  —  16 
■  =17;  and  the  number  is  divisible.     If  the  given  number  were 
not  even,  make  it  so,  by  the  addition  or  subtraction  of  17. 

In  1,581,  for  example,  add  17  to  the  last  two  figures ;  you  get 
98,  one  half  of  which  is  49;  and,  49  — 15  being  34,  the  number 
is  divisible  by  17. 

N.  B. — In  some  cases  it  will  be  more  simple  to  double  the  first 
part  of  the  number.  For  example,  in  1,581,  you  may  double  15, 
and  subtract  30  from  81,  which  is  51  =  3  X  17. 

3.  19  divides  a  number  when  twice  its  last  figure^ 
added  to  the  other  "party  is  a  multiple  of  19 » 

114  is  divisible  by  19,  because  2  X  4+ 11  =  19. 
798  "  "    because  2  X  8  +  79  =  95  =  5  X  19. 

95         itself,  because  2  X  5+ 9  =  19. 

I  do  not  give  demonstrations  of  all  these  properties  of 
numbers ;  they  would  appear  too  abstruse  at  present,  and 
in  fact  belong  rather  to  algebra. 

I  have  added  to  the  rules  frequently  met  with  in  arith- 
metics, those  for  7,  13,  17,  and  19,  which  I  deem  suffi- 
ciently simple  to  be  performed  mentally.  Beyond  19, 
the  rules  become  too  complicated  for  practical  purposes. 

Questions, — What  are  the  rules  of  divisibility  by  11  ?  12?  13? 
17?  19? 

EXERCISES. 

1.  Which  of  the  numbers  104;  695;  9,099;  615,602;  304,672; 
92,416,493;  21,859,838;  2,649,257,  are  divisible  by  13? 

2.  Which  of  189;  266;  2,394;  4,368;  7,728,  are  divisible  by  63; 
and  which  by  14?     (XXXIL,  1.) 

3.  Which  of  225;  540;  1,296;  1,485;  396,  are  divisible  by  15? 
36?  44?  99?  55?  495?     (XXXVI.,  1.) 

4.  Which  of  255;  363;  4,284;  6,001;  724;  816,  are  divisible 
by  17? 

11* 


126  LESSON  xxxvr. 

5.  Which  of  171;  215;  361;  589;  617;  931;  1,026,  are  divisi- 
ble by  19? 

6.  How  many  times  is  2  a  factor  in    8;  in  46;  in  64? 

"  3         "  27;  in  99;  in  783  ? 

"  7         "  35;  in  49;  in  84;  in  343? 

7.  What  are  the  simple  factors  of  324;  640;  826;  375;  2,625; 
588;  207,025;  20,025;  8,119;  32,476;  357,236;  224,939? 

LESSON  XXXVI. 

This  lesson  will  contain  some  general  remarks  concern- 
ing the  divisibility  of  numbers,  which  will  facilitate  the 
application. of  the  foregoing  rules. 

1.  A  number  is  exactly  divisible  by  a  multiple  when 
it  contains  all  its  component  factors. 

Thus,  12  divides  192,  because  it  is  divisible  by  the  component 
factors,  3  and  4,  of  12;  and  we  have  seen  (XXXII.,  1)  that  we 
may  divide  by  the  factors  instead  of  the  whole  divisor. 

2.  Every  number  which  divides  another,  divides  any 
one  of  its  multiples. 

In  other  words,  a  number  which  divides  one  factor^ 
divides  the  product. 

Exr — Since  4  divides  12,  it  divides  also  192. 

This  follows,  also,  from  Lessons  XXXII.  and  XXXITI. 
We  may  get  4  out  of  each  12  by  three  subtractions,  and 
thus  take  out  successively  the  sixteen  12s  contained  in 
192. 

3.  A  number,,  prime  to  others^  cannot  divide  their 
pi'oduct, 

60  is  equal  to  6  X  10,  and  5  divides  it,  because  it  divides  10; 
but  it  does  not  divide  63,  because  it  divides  neither  7  nor  9. 

■  For,  by  reference  to  the  table  of  Pythagoras,  it  will 
appear  evident  that  the  repetition  of  7  can  make  the  same 
number  as  the  repetition  of  5,  only  where  the  horizontal 
column  of  the  one  meets  the  vertical  column  of  the  other ; 
whereas,  the  multiple  63  is  at  the  meeting  of  the  column 
7  and  9. 


DIVISIBILITY    OF    NUMBERS.  127 

4f.  A  number  which  divides  exactly  each  part,  divides 
also  their  sum, 

36  is  equal  to  12  +  24 ;  both  of  which  are  divisible  by  4  as  well 
as  36. 

For,  we  may  take  out  the  first  part  by  successive  sub- 
tractions J  then  the  second  part  will  remain,  and  may  be 
divided  exactly. 

5.  A  number  which  divides  another  and  one  of  its 
parts,  divides  also  the  other  part. 

Or,  in  other  words,  a  number  which  divides  two  others, 
divides  their  difference, 

4  divides  36  and  12;  consequently,  also,  36 —  12  =  24. 

For,  we  can  take  out  the  first  given  part  by  successive 
subtractions  of  the  divisor;  and  if,  after  continuing  the 
subtractions,  the  last  part  left  a  remainder,  it  would  evi- 
dently follow  that  the  whole  was  not  divisible,  as  admit- 
ted in  the  premises. 

6.  From  the  two  preceding  propositions,  we  conclude 
that: 

When  you  wish  to  test  the  diinsibility  of  a  number  by 
a  factor,  you  need  not  consider  any  figure  or  part  of  the 
number  which  is  evidently  divisible  by  the  given  factor. 

If  it  were  required  to  ascertain  whether  714,273  is  divisible  by 
7,  we  might  omit  the  two  7s  and  14 ;  so  that  the  trial  would 
depend  only  on  203,  and  because  2  X  3  =  6  and  20  —  6  =  14,  the 
divisibility  would  be  proved  (XXXIV.,  6). 

Again:  take  264,654,  to  be  divided  by  13:  we  may  take  out 
26  and  65,  which  are  multiples  of  13,  and  try  only  4,004;  which 
is  divisible  by  13,  because  the  two  periods  are  equal. 

Here,  the  parts  being  separately  divisible,  the  whole  is  so  like- 
wise (4). 

7.  By  the  application  of  the  same  principle,  we  may 
also  frequently  ascertain  lohether  a  number  is  divisible 
or  not  by  another  (even  a  large  one),  by  adding  or  sub' 
trading,  at  any  place  of  it,  such  a  convenient  auxiliary 
number  as  will  make  it  divisible  by  the  given  factor. 

Then,  if  this  auxiliary  number  is  not  divisible,  the 


128  LESSON   XXXVI. 

whole  mimber  itself  is  not  so  ;  if  it  is,  the  whole  number 
is  divisible. 

343  is  divisible  by  7,  because  the  addition  of  7  makes  it  350, 
which  is  7  times  50.     Hence,  343  =  7  X  49. 

241  is  not  divisible  by  7,  because  you  must  take  from  it  the 
indivisible  number,  31,  to  make  it  the  multiple,  210,  of  7. 

13  does  not  divide  255,  because  it  takes  only  5  more  units  to 
make  it  the  multiple  260  =  13  X  20;  so  that  255=  13  X  19  +  8. 

But  13  divides  234,  because  the  addition  of  the  multiple,  26, 
makes  it  260.     Hence,  234  =  13  X  18. 

Even  the  large  prime  number,  367,  may  be  readily  discovered 
to  be  a  factor  of  3,303,  and  also  of  4,037 ;  because  the  addition  of 
367  to  the  first,  makes  it  3,670  =  367  X  10  :  so  that  3,303  = 
9  X367. 

As  regards  the  second  number,  very  little  practice  in  numbers 
will  enable  any  one  to  see  at  a  glance  that,  if  367  be  subtracted 
from  the  first  part,  403,  the  remainder  would  be  36  tens,  which, 
joined  to  the  last  figure,  makes  it  367.  Hence,  4,037  =  11  X  367, 
being  the  sum  of  the  tw^o  multiples  3670  and  367. 

The  number  30,927  will  be  discovered  to  be  divisible  by  13, 
more  promptly,  by  taking  away  91  from  the  middle  part,  92; 
when,  in  the  remainder,  30,017,  it  will  be  at  once  discovered 
that  the  difference  of  the  periods  is  13  (XXXV.,  1). 

I  merely  point  out  this  manner  of  discovering  the  fac- 
tors of  numbers,  which,  with  some  practice,  is  applicable 
in  a  great  many  cases ;  but  no  precise  rule  can  be  given. 
The  successful  application  of  the  method  depends  alto- 
gether on  the  degree  of  readiness  and  dexterity  of  each 
individual.  The  young  arithmetician  should  exercise 
himself  in  applying  it,  and  it  will  not  be  long  before  he 
has  acquired  much  expertness  in  it.  This  will  greatly 
improve  his  facility  in  calculations.* 

*  The  trials  to  discover  the  prime  factors  of  a  number,  need  not 
be  carried  beyond  that  number  (square  root)  which,  multiplied  by 
itself,  would  give  the  nearest  product  of  equal  factors  below  the 
given  number. 

For  example:  we  wish  to  find  whether  1,373  has  any  factors;  we 
try  successively  every  prime  number  until  we  come  to  37;  and 
observing  that  37  X37=  1,369,  and  that  the  next  prime  number,  41, 
multiplied  by  itself,  will  exceed  1,600,  and,  consequently,  our  given 
number,  we  stop  the  trial  at  37;  because,  evidently,  any  number 
above  it  would  require  to  be  multiplied  by  a  factor  smaller  than  37, 
and  therefore  already  tried.  Hence,  we  conclude  that  1,373  is  a 
prime  number. 


DIVISIBILITY    OF    NUMBERS.  129 

Questions, — If  the  component  factors  divide  a  number,  will 
the  multiple  divide  it  also  ?  If  a  number  divides  one  factor,  does 
it  divide  the  product?  Why?  When  may  a  number  divide  the 
product  of  two  others  ?  When  w^U  it  not  ?  If  two  numbers 
nave  a  common  factor,  will  it  divide  their  sum  and  difference  ? 
In  testing  the  divisibility  of  a  number,  what  parts  of  it  may  be 
neglected  ?  How  may  an  auxiliary  number  be  used  to  simplify 
the  test  of  divisibility  ? 


END  OF  PART  I. 


d 


PART  II. 


The  First  Part  contains  all  the  principles  and  rules 
necessary  to  understand  and  execute  with  facility  every 
elementary  operation  in  arithmetic.  Their  applications 
to  practical  questions,  which  require  generally  a  combi- 
nation of  several  of  the  rules,  will  form  the  subject  of  the 
second  part. 

When  a  question  is  proposed,  the  first  step,  which  is 
commonly  the  most  embarrassing  part  of  the  process  of 
solution,  consists  in  reducing,  by  a  course  of  reasoning  and 
analysis,  the  conditions  of  the  question  to  a  numerical 
statement,  indicating,  by  means  of  the  usual  signs,  the 
various  elementary  operations  to  be  performed. 

The  next  step  is  more  mechanical,  and  consists  merely 
in  working  out  the  result  from  the  numerical  statement, 
by  means  of  the  elementary  rules  laid  down  in  the  first 
part. 

The  analysis  of  practical  questions,  as  well  as  the 
arrangement  and  simplifications  of  numerical  {statements, 
constitute  the  chief  object  of  the  Second  Part ;  in  which 
no  new  operation  will  be  found,  but  only  applications  of 
those  already  known. 

For  the  final  execution  of  the  operations,  reference 
will  be  made  to  the  first  part. 

I  would  advise  the  student  never  to  omit,  and  the 
teacher  always  to  require,  the  setting  down  at  the  head 
of  the  operations  the  numerical  statement  of  each  ques- 
tion, in  the  manner  that  will  be  shown  in  the  various 
examples  I  shall  have  occasion  to  explain.  There  are 
great  advantages  in  this  method ;  it  requires  and  shows  a 
correct  investigation  of  the  data  j  it  enables  the  operator 


TRANSFORMATION   OF   FRACTIONS.  131 

to  discover  cancellings  and  other  simplifications,  which 
the  combination  of  operations  frequently  allows  of;  and, 
finally,  in  case  of  some  error  having  been  committed,  it 
facilitates  its  detection  and  the  revision  of  the  whole 
process. 

By  being  careful  in  the  beginning  to  arrange  his  work 
systematically  and  neatly,  the  student  will  soon  acquire 
the  faculty  to  embrace  in  his  mind  the  numerical  state- 
ments and  the  simplifications  they  lead  to ;  and  thereby 
he  will  obtain  great  mental  readiness  in  calculations. 


CHAPTER  VI. 


CONTAINING    OPERATIONS    IN   VULGAR    FRACTIONS   AND 
PROPORTIONS. 

LESSON  XXXVII. 

TRANSFORMATION  OF  FRACTIONS  AND  NUMBERS,  COMMONLY  CALLED 

REDUCTION. 

I  HAVE  placed  the  complement  of  the  doctrine  of  frac- 
tions in  the  Second  Part,  because  what  follows  involves 
no  new  principle ;  and  each  question,  when  properly 
investigated,  is  a  mere  repetition  or  application  of  what 
precedes.  The  operations  on  fractions  belong,  therefore, 
to  the  practical  part  of  the  work ;  and,  like  the  rest, 
should  not  be  entered  upon  until  the  theoretical  princi- 
ples and  elementary  rules  have  been  completely  mastered. 


1.  To  transform  (reduce)  an  improper  fraction  into  its 
equivalent  whole  or  mixed  number. 

In  conformity  to  a  general  practice,  I  here  make  use 
of  the  word  improper  fraction,  and  introduce  the  present 
rule,  though  the  case  is  nothing  more  than  common  divi- 
sion. 


132  LESSON   XXXVII. 

Let  us  take,  for  example,  as  an  improper  fraction,  ^^. 
What  is  it,  if  not  12  divided  by  4  =  3  ? 

Again  :  vi^hy  should  \^^  be  called  an  improper  fraction ; 
and  what  need  is  there  of  a  new  rule  to  find  out  that 

751  divided  by  17,  is  equal  to  44jV  • 

The  rule  for  division  has  taught  hov^  to  obtain  the  in- 
tegral part,  44,  of  the  quotient,  as  well  as  its  fractional 
part,  j\,  by  setting  the  remainder  over  the  divisor. 

2.  I  consider  it,  therefore,  almost  useless  to  state  that, 
to  transform  an  improper  fraction  into  its  equivalent 
whole  or  mixed  number^  you  divide  the  numerator  by  the 
denominator^  as  in  common  division. 


3.  To  transform  a  whole  or  mixed  number  into  its 
equivalent  improper  fraction. 

We  have  seen  (XXX.,  8)  how  a  whole  number  may  be 
put  under  the  form  of  a  fraction,  with  a  given  denomi- 
nator. 

In  the  case  of  mixed  numbers,  therefore,  all  we  have 
to  do  is  to  put  the  whole  number  under  a  fractional  form, 
having  the  same  denominator  as  the  fraction,  and  then  to 
add  the  numerators.     Let  us  take,  for  example, 

44jV 

We  know  that  44  =  — yj —  =  -j= ,  and  this  added  to 

3_  .    751 

17'  ^^  17  • 

So  that  this  is  not  a  new  operation. 

4.  Multiply  the  whole  number  by  the  denominator  of 
the  fraction ;  to  the  product  add  the  numerator  of  the 
fraction,  and  place  the  result  over  the  said  denominator^ 

This  is  evidently  the  reverse  of  the  preceding  case, 
and  requires  exactly  the  same  operations  as  the  proof  of 
division  (XVIIL,  5).  For,  since  751  divided  by  17,  is 
44j^y,  if  we  wished  to  go  back  to  the  dividend,    we 


TRANSFORMATION   OF    FRACTIONS.  133 

would  multiply  the  divisor,  17,  by  the  quotient,  44,  and 
add  the  remainder,  3. 

N.  B. — This  rule  may  also  be  understood  by  consider- 
ing that  1  unit  is  17  seventeenths,  and,  consequently,  44 
is  equal  to  44  times  this  j  that  is,  44  X  17  =  748  seven- 
teenths, and  that  3  seventeenths  more  make  751  seven- 
teenths or  Yt  .  But  the  above  mode  of  reasoning,  in  this 
case,  is  more  systematical,  inasmuch  as  it  retains  the  same 
mathematical  uniformity  of  investigation. 

CASE    III. 

5.  To  reduce  a  fraction  to  its  simplest  terms,  apply 
the  rules  for  the  divisibility  of  numbers,  and  divide  both 
terms  by  all  the  common  factors  you  may  discover 
(XXX.,  9). 

Or,  if  the  numbers  are  large,  find  their  greatest  com-- 
mon  divisor,  and  divide  both  terms  by  it. 

The  first  method,  with  a  competent  knowledge  of  the 
divisibility  of  numbers,  will  generally  suffice,  and  prove 
the  shortest. 

6.  Sometimes,  however,  a  large  divisor,  common  to 
both,  may  not  be  apparent.  In  that  case,  resort  must  be 
had  to  the  method  of  the  greatest  common  divisor,  which 
will  be  the  subject  of  the  next  lesson. 

Questions. — How  is  an  improper  fraction  transformed  into  its 
equivalent  whole  or  mixed  number  ?  How  is  a  whole  or  mixed 
number  put  under  a  fractional  form?  What  simple  rules  does 
each  operation  correspond  to  ?  How  is  a  fraction  reduced  to  its 
lowest  terms  ? 

EXERCISES. 

1.  iransiorm  y  ;  jq-  ,  ~25~;  T^T";  lUT  ;  iTlJ  >  ^^ 
equivalent  whole  or  mixed  numbers. 

2.  Changeintofractions,  12|;  12^5  14/^;  151f;  ISS/^ 
=  ^If^j  178214  =  ^-^i-p  ;  6579^Vi/o- 

3.  Reduce  to  their  simplest  terms,  by  cancelling  factors, 
12 


lU 


LESSON   XXXVIII. 


48.     60._8  4_.23.8  —  17.1344  7.  2^0  5  .     192. 

■JS"  >  T2  5  ?    1  7  0  ?    2'5'2   —    1  ^  J    idle'  —  ¥  7  "25H  —  6'?    ^4  0  ' 

175  0   5  .       ^8,2  2  0     13 

3"T5  (J  —  ¥  >     17  6".4U(J  —  6  0* 

4.    Change  15  into  a  fractional  expression  whose 

denominator  shall  be  .         .  7 

23  «  <«  10 

45  «  «  25 

51  «  «  500 

65  «  «  272 


I; 


LESSON  XXXVIII. 
THE  GREATEST  COMMON  DIVISOR. 

1.  The  operation  for  findino;  the  greatest  common  divi- 
sor,  is  a  series  of  divisions.  It  depends  on  the  principle 
explained  in  Lesson  XXXVL,  5,  that, 

A  number  idMcJi  divides  another  and  one  of  its  parts, 
divides  also  the  other  j)art, 

2.  Let  it  be  supposed,  for  instance,  that,  in  order  to 
reduce  the  fraction 

2.76 
3 '60" 

to  its  lowest  terms,  we  wish  to  find  the  greatest  common 
divisor  between  its  numerator  and  denominator :  we  re- 
mark, first,  that  this  number  cannot  be  greater  than  276  ; 
and,  therefore,  it  is  proper 


OPERATION. 


360 

84 


1 


276 
24 


2 
12  =  Ans, 


to  try  whether  276  itself 
might  not  be  the  greatest 
common  divisor.  W'iththis 
view, we  divide  360  by  276, 
and  get  a  quotient,  1,  which 
we  set  above  the  divisor,  276. 

Because  there  is  a  remainder,  84,  we  conclude  that  276 
is  not  the  greatest  comfnon  divisor. 

But,  whatever  this  greatest  common  divisor  may  be, 
since  it  must  divide  360  and  the  part,  276,  now  taken  out 
of  it,  it  must  also  divide  the  other  part,  84  (XXXVL,  5). 

The  question  is,  therefore,  reduced  to  finding  the 
greatest  common  divisor  between  276  and  84. 


THir  GREATEST    COMMON    DIVISOR.  ISfll 

For  this  purpose,  upon  the  same  principle  as  before, 
we  divide  276  by  84,  and  get  a  quotient,  3,  which  we 
set  likewise  above  the  new  divisor. 

There  being  still  a  remainder,  24,  we  conclude  that 
84  is  not  yet  the  greatest  common  divisor  sought. 

But,  whatever  it  may  be,  we  know  that  it  divides  84 
and  276,  and  therefore,  also,  the  new  remainder-  and  the 
question  is  further  reduced  to  finding  the  greatest  common 
divisor  between  84  and  24. 

By  a  new  division,  we  have  now  a  quotient,  3,  and 
another  remainder,  12:  which  shows  that  24  itself  is  not 
the  number  sought,  and  that  we  must  look  for  it  in  the 
last  remainder,  12. 

Accordingly,  we  divide  24  by  12;  and,  getting  an 
exact  quotient,  2,  we  conclude  that  12  ts  the  gi-eatest 
common  divisor, 

3.  For,  it  divides  24  and  all  the  preceding  remainders, 
which  are  composed  of  multiples  of  24  and  12,  as  can 
easily  be  verified. 

24  =  12  X  2. 

84=24X3  +  12. 
276=  84X3  +  24=  23X12. 
3G0  =  276  +  84  =  30  X  12. 

4.  Furthermore,  it  is  evidently  the  greatest,  because  12 
is  one  of  the  remainders,  which  must  contain  it. 

We  conclude,  therefore,  that  the  simplest  form  of  the 
fraction  f^,  is  ||. 

The  example  shov^^s  the  best  manner  of  arranging  the  opera- 
tion. 

5.  Hence,  to  find  the  greatest  common  divisor  between 
ixoo  numbers : 

I.  Divide  the  greater  number  by  the  less  ;  this  by  the 
remainder  ;  then,  the  first  remainder  by  the  second  ;  the 
second  by  the  third^  and  so  on,  until  one  of  them  divides 
exactly  the  'pi^eceding.  This  last  remainder  will  be  the 
greatest  common  divisor. 


136 


LESSON   XXXVIII. 


.  II.  If  the  last  remainder  he  1,  we  conclude  that  the 
numbers  have  no  other  common  measure  than  unity  ;  or, 
in  other  words,  that  they  are  prime  to  each  other ^  and 
the  fraction  irreducible. 

As  regards  the  last  case,  let  us  take  the  fraction  ffj. 


OPERATION. 


2 

1 

3 

15 

1 

1 

2 

873 

317 

239 

78 

5 

3 

2 

1 

239 

78 

5 

3 

2 

1 

0 

The  operation  shows  that  no  other  number  than  unity 
divides  both  terms :  consequently,  the  iraction  is  in  its 
simplest  form,  and  there  is  properly  no  common  divisor. 

6.  It  furnishes  also  another  useful  remark.  When,  in 
the  course  of  the  operation,  one  of  the  remainders  is  a 
prime  number,  which  does  not  divide  the  preceding 
remainder,  it  is  certain  that  no  common  divisor  exists, 
and  it  is  unnecessary  to  continue  the  operation. 

In  the  last  example,  5  does  not  divide  78;  and,  consequently, 
since  it  is  a  prime  number,  no  divisor  can  be  found  for  it,  but 
unity,  and  it  v^'ould  be  wrong  to  carry  the  operation  further. 

7.  To  find  the  greatest  common  divisor  betiveen  several 
numbers,  find  it  between  the  two  smallest  of  them-,  then, 
between  this  and  the  third  number  in  size  ;  between  the 
second  divisor,  thus  obtained,  and  a  fourth  number  ;  and 
so  on,  to  the  last. 

It  is  evident  that  the  greatest  common  divisor  between 
several  numbers,  cannot  exceed  the  common  divisor  be- 
tween any  two  of  them. 

Therefore,  if  we  knew  them  all  for  pairs,  there  would 
only  remain  to  find  the  number  common  to  them  all. 

For  this  reason,  to  proceed  systematically,  we  begin 
with  the  two  smallest  numbers,  and  continue  as  above 
stated. 


THE    GREATEST    COMMON    DIVISOR. 


137 


Let  us  take,  as  an  example,  504,  756,  1260,  and  2,058. 


In  the  firstplace,we  perform  the 
operation  between  504  and  756, 
and  find  that  their  greatest  com- 
mon divisor  is  252.  We  know 
that  this  particular  greatest  divi- 
sor must  contain  the  general  one. 

Therefore,  we  now  operate  upon 
it  and  1260,  and  find  that  252  an- 
swers also  for  1260. 


FIRST  OPERATION. 

1   |2 
756  5041252 
252 


SECOND  OPERATION. 

5 


1260 
•00 


252 


Lastly,  we  operate  with  2058,  and 
find  42  for  the  final  result. 


THIRD  OPERATION. 

6 
42 


2058 
42 

8 
252 

REMARKS. 

The  operation  of  the  greatest  common  divisor  should 
always  be  simplified,  when  practicable,  as  follows : 

8.  Remark  I. — Strike  out  any  prime  factor  contained 
in  one  of  the  numbers,  and  not  in  the  other.  For,  it  is 
evident  that  it  can  make  no  part  of  a  common  divisor. 

For  example,  if  we  take  249  and  377 : 

We  observe  that  249  is  divisible  by  3 ;  while  377  is  not.  Con- 
sequently, 3  can  be  no  part  of  the  greatest  divisor.  I,  therefore, 
suppress  it  out  of  249,  which  leaves  83 ;  which,  being  a  prime 
number,  leads  me  at  once  to  the  conclusion  that  83  must  be  the 
greatest  common  divisor,  if  any. 

But,  377  is  not  divisible  by  83;  therefore,  without  further 
operation,  it  is  made  apparent  that  ^249  and  377  are  prime  to  each 
other,  and  ^4^  irreducible. 

'  3  7  7 

9.  Remark  II. —  Tahe  out  any  factor  common  to  both, 
and  set  it  aside  as  part  of  the  greatest  common  divisor^ 
hy  which  to  multiply  the  final  result  of  the  operation. 

For  example,  in  2,150  and  3,612,  I  see  at  once  that  25  is  a 
factor  of  the  first  number,  but  not  of  the  second ;  and,  according 

12* 


I3S 


LESSON   XXXVIII. 


to  the  first  remark,  I  strike  it  out ;  by  which  the  trial  is  now  to 
be  made  on  the  smaller  number,  86,  with  3,612. 

Now,  I  see  that  2  divides  both ;  hence,  2  is  part  of  the  greatest 
common  divisor.  Dividing  86  by  it,  the  quotient  is  43 ;  and, 
because  this  is  a  prime  number,  it  must  be  the  reduced  greatest 
common  divisor,  if  any.  A  trial  proves  it  to  be  so ;  consequently, 
the  whole  divisor  is  2  X  43,  and  the  first  number,  2,150,  is  equal 
to  86  X  25 ;  the  second,  to  86  X  42. 

Questions, — What  is  the  greatest  common  divisor  ?  Give  the 
rule  for  finding  it.  What  principle  does  the  rule  depend  on? 
How  is  the  operation  arranged  ?  Where  do  you  put  the  quotient  ? 
How  do  you  know  that  there  is  no  common  divisor?  What  is 
the  character  of  the  fraction  in  that  case  ?  If  you  come  to  a 
prime  number  for  a  remainder,  and  it  does  not  divide  the  pre- 
ceding one,  what  do  you  conclude  ?  How  is  the  greatest  common 
divisor  between  several  numbers  found  ?  Why  ?  What  may  be 
4one,  if  one  number  contains  a  factor  not  in  the  other  ?  If  it 
contains  one  common  to  both  ? 


EXERCISES. 

Find  the  greatest  common  divisor  of 

1.  246  and  372. 

2.  9,024  and  3,760. 

3.  637  and  143. 

4.  999  and  592. 

5.  3,072  and  912.   . 

6.  138,174  and  49,365. 

7.  132,568  and  920,712. 

8.  16,784  and  3,144. 

9.  101,549  and  40,103. 


Ans.   6. 
Ans,   752. 
Ans,   13. 
A71S,   37. 
A71S.   48. 
Ans,   3. 
Ans.   1,816. 
Ans.   8. 
Ans.   7. 


Reduce  to  their  lowest  terms,  by  the  greatest  common  divisor, 


1. 

"6T6-* 

Ans. 

2. 

252 

•3  6:t* 

Ans. 

3. 

mt' 

Ans. 

4. 

4VA- 

Ans. 

5. 

5Q65 

5T-2  0-- 

Ans. 

6. 

7  9  44 

Ans. 

1 

9 

7 

J 

5* 
1  0  ?, 
1  04* 
6  62 
T4  3-* 


7. 

8. 

9. 
10. 
11. 
12. 


8  9jl4 
44  9  4' 
24  3  3 

314  17.5 
1  00005* 

1  1  "5  4  4  • 

1  6_6_9_3_3 

i'  -J  5"  5  7  • 

10  0  110 

3"Ts(r6"  • 


Ans. 
Ans. 
Ans. 
Ans. 
Ans. 
Alts. 


212 
1  (J7' 

3 

T7* 

Z'^  "^ 
TT3' 

9 

IT* 

329 
4  I  • 

3  5  T 
TTT* 


TRANSFORMATION  OF  FRACTIONS.  13? 

1 A       JilhJLl  i'5J7_4, 9  9  4^.43  2     16,807 

•'^*      1-19,7  20,2  3  7, 9  27, 424'  —   5  9,04T' 

l/T       7,241,379,310,344,827,5  86,206,896,551   21 

^^*     V,'9[)  9,y  9S",9  9ir,9  9  9,'9T9,"9F9,"S"9T,'9T9,"99T  —   2  9* 

1.  What  is  the  greatest  common  divisor  between  336;  720;  1736? 

A71S,     8. 

2.  «  «  2961;  799;  564;  94?        A7is.  47. 

3.  «  «  210;  462;  2157;  189'/      Atis.  21. 


LESSON  XXXIX. 

TRANSFORMATION  OF  VULGAR  INTO  DECIMAL 
FRACTIONS. 

1.  A  great  many  operations  are  much  simplified  by  this 
transformation.  A  common  fraction,  it  has  been  said,  is 
nothing  but  the  expression  of  the  quotient  of  the  nume- 
rator by  the  denominator. 

2.  Hence,  to  change  a  vulgar  into  a  decimal  fraction : 

Divide  the  numerator  by  the  denominator^  annexing 
ciphers  to  the  numerator  as  far  as  necessary^  to  give  the 
requisite  number  of  decimals. — For  example, 

3.  Let  it  be  proposed  to  transform  |-  into  a  decimal 
fraction.  Since  7  units  do  not  contain 
8,  there  are  no  units  in  the  quotient, 
and  we  change  the  7  into  70  tenths^  by 
adding  a  0  to  it,  and  continue  the  divi- 
sion by  adding  Os  to  the  remainders,  as 
in  the  case  of  the  extension  of  the  quo- 
tient, (XXVIL,  3.) 

4.  Two  distinct  cases  will  occur  in  this  transformation. 

I.  The  division  may  terminate,  and  an  exact  quotient 
be  obtained. 

II.  The  division  may  never  end:  then  the  quotient 
will  be  an  infinite  decimal. 


OFERATION. 

70      18 


5.   The  division  will  always  end  wher&the  denomina- 
tor does  not  contain  any  other  factor  than  2  and  5. 


140  LESSON    XXXIX. 

For,  it  is  readily  seen  that  either  2  or  5  will  divide 
exactly  any  number  as  many  times  as  there  are  Os  at  the 
end  of  it.     Thus : 

70     will  be  divided  once         by  either    2  or  5. 
700  «  twice  «  " 

7,000  «  three  times      «  " 

The  last  for  example,  divided  three  times  by  2,  might  be 
written 

7_><_10_X J.0  XJ.0 
"~2~X~2~><~2~~; 

where,  evidently,  each  2  may  be  cancelled  out  of  the  corre- 
sponding 10.     And  the  same  thing  may  be  said  of  5,  in 

7X10X10X10 
5X5X5      • 

Therefore,  I,  because  2  is  three  times  factor  in  8, 
will  require  the  addition  of  three  Os  to  the  numerator, 
changing  the  fraction  into  ^-g-^,  the  quotient  of  which 
will  end  at  the  third  decimal  place  ;  and  we  get  |  =  0.875. 

i?,  in  which  5  is  twice  factor,  will  require  the  addition  of 
onfy^two  zeros,  and  the  exact  quotient  will  have  but  two  deci- 
mals, and  we  get  -J  5  =  0.52. 

^317        ,  .  ,   .  ,  31 

In  the  fraction    ^-g^,  which  is  equal  to  2X5X5X5X5' 

the  factor,  5,  being  found  four  times  in  the  denominator,  the 
complete  quotient  will  contain  4  decimals,  and  we  shall  find 
A'ra  =  0-2536. 

6.  In  general,  the  quotient  icill  contain  a  number  of 
figures  equal  to  the  greatest  number  of  times  that  either 
2  or  5  is  factor  in  the  divisor  (that  is,  to  the  greatest 
power  of  'either  3  XXXIIL,  11). 

For,  it  is  only  the  factor  which  is  found  the  greatest  number 
of  times  in  the  denominator,  that  determines  the  number  of  deci- 
mal places ;  so  that,  in  the  last  example,  we  have  four,  and  not 
five,  of  them.  The  reason  of  this  is,  that  each  10  is  formed  of  5 
and  2  as  joint  factors ;  therefore,  the  exact  division  by  5  implies 
that  by  2. 

The  above^emark  will  enable  the  operator  to  deter- 


TRANSFORMATION   OF    FRACTIONS.  l^^* 

mine  how  far  the  division  is  to  be  carried,  if  he  wishes 
to  get  a  complete  quotient :  it  will  also  be  a  check 
aojainst  mistakes. 


CASE   II. 

7.  The  division  cannot  end  when  any  other  prime  fac* 
tor  than  2  or  b  exists  in  the  denominator. 

For,  no  other  prime  factor  divides  exactly  10;  and, 
consequently,  none  can  divide  its  decimal  multiples  or 
powers  100,  1000,  &c.  (XXXVL,  3.) 

In  this  case,  therefore,  the  decimal  fraction  will  be  in- 
finite ;  and,  furthermore,  if  carried  far  enough,  the  same 
figures  will  be  repeated  periodically.  For,  since  the 
quotient  cannot  end,  we  must  at  last  exhaust  all  the  forms 
of  the  remainder,  and  come  back  to  one  of  those  that 
have  preceded,  when  the  same  rotation  of  quotients  wil] 
return.     Thus  we  find : 

I  =  0.3333,  &c.  =  0.3.  f    =  0.857142857142,  &c. 

y\  =  0.272727,  &c.  =0.27.  =  0.857142. 

\\  =  1.090909,  &c.  =  1.09.  j-|-  =  0.91666,  &c.  =  0.9ia! 

ij-  =  0.351351,  &c.  ==  0.351.  /i-  =  0.201666,  &c.  =  0.2916^ 

These  infinite  decimal  fractions  are  called  repeating  or 
periodical  fractions. 

N.  B. — To  show  that  the  figures  must  be  conceived  to  be  inde- 
finitely repeated,  a  conventional  sign  is  made  to  separate  the  pe- 
riod. Some  use  a  dot  over  each  figure  of  the  period ;  but  it  is 
not  as  distinct  as  the  angular  line  (i )  here  exhibited. 

8.  If  the  denominator  of  the  fractions  produces  a  long 
period,  and  it  is  not  important  to  carry  the  operation  so 
far,  the  sign  +  may  be  placed  after  the  last  decimal 
figure  (XXVII.,  4.) 

Thus,  we  may  write  y  =  0.857  +,  if  we  do  not  wish  to  obtain 
a  nearer  approximation  than  one  thousandth. 


I 


142  LESSON   XXXIX. 

9.  This  sign  is  omitted  when  the  decimal  part  is  judged 
to  be  sufficiently  accurate  for  the  purpose  intended ;  then 
it  is  customary  to  add  1  to  the  last  figure,  if  the  following 
is  5  or  more  (XXVII.,  note.)     Thus,  we  would  write : 

I  =0.67;  y\=  0.273;  f=0.86; 

-}-^-=  0.917,  or  =0.9167,  )  according   to   the   degree  of 
_7_=,    .2917,010.29167,1        approximation  wanted. 
It  may  be  observed  that  sometimes,  as  in  the  case  of 
is  J  /4J  the  period  does  not  begin  at  once. 

10.  The  period  generally  begins  after  a  number  of 

decimals  equal  to  the  greatest  number  of  times  that  either 

2  or  5  (not  both)  is  factor  in  the  deiiominator . 

4  I — 

Thus,  yV  = =  0-053;  the  period  begins  after  two 

«^  X  o  X  5  figures. 

24  =  3v2y2v2  =  ^'^^^'^  ^^^  P''^°^  ^"S^"'  ^^" 
o  p^  ^  X  ^  A  ^  ter  three  figures. 

11.  This  can  be  readily  demonstrated  by  multiplying 
both  terms  of  the  fraction  by  as  many  equal  factors  as 
will  change  each  complement  factor  of  its  denominator 
into  10. 

Thus,  if  we  had  to  transform  ^^  5?  which  is  equal  to  ^      ^      ^ 

y  X  «^  X  «^j 
the  multiplication  of  both  terms  by  twice  2,  would  change  it  into 

2x2x2  8  » 

9  X  10  X  10  ^  9^100-    N°^'  f  ^  ^-S^'  ^^^  evidently  the 
^f  division  by  100  would  push  the  period  two  places  to  the  right, 

J    I  and  make  it  0.0088,  which  begins  at  the  third  figure. 
^   TRANSFORMATION   OF  DECIMALS  INTO   VULGAR 
FRACTIONS. 

?  12.  Tlace  the  decimal  fraction^  as  a  numerator^  over 
the  number  which  gives  its  name  to  its  last  units;  or, 
in  other  words,  over  1,  loith  as  rnany  O5  as  there  are  de^ 
cimal  figures. 

Thus,  0.2;     0.25;    0.009;    4.19;  5.0007,  will  be,  respec- 

lively,      y  (J ,     1  oTJ '   ToiyXJ '  ^ i  o^  "^  T7JTJ '  *^To,o  q-o  ^^  To,oFo • 

Then,  if  practicable,  you  may  reduce  the  fraction  to  its  lowest 
terms,  as  in  the  first  two  examples,  which  become  -5  and  \* 


TRANSFORMATION   OF   FRACTIONS.  143 

13.  In  the  case  of  a  rejpeatmg  fraction^  we  moAj  return 
to  the  original  fractio?i  by  lolaciyig  the  jperiod  over  as 
many  Qs  as  there  are  decimals^  and  then  reduciiig  tJie 
fraction  down  to  its  lowest  terms  (XXXVII.,  5.) 
Thus :  __ 

0.3^^  =  ^;  0.27^=J^  =  ^3_  .    o:9  =  |.=  l; 

0.351  =r  l|i  =  |A ;    0.857142  =  |||^|  =  f 
This  is  easily  demonstrated  by  considering  that : 

i  =  0.111,  &c.;  -yV  =  0.0101,  &c.;  y^|^  =  0.001001,  &c. ; 
or,  in  general,  that  1  divided  by  a  number  composed  of  9s,  forms 
a  period  of  as  many  figures  as  there  are  9s ;  this  period  consist- 
ing of  1  preceded  by  Os.  Now,  any  other  period  may  be  consi- 
dered as  the  product  of  the  number  which  forms  it,  by  one  of 
the  above.     For  example  : 

0.351351,  &c.  =  351  X  0.001001,  &c.  =  ||^  =  jf . 
If  the  'period  does  not  begin  next  to  the  units'*  points 
transform  it  firsts  reduce  and  annex  the  fraction  to  the 
decimal  part  'which  precedes ;  then  change  the  mixed 
number^  thus  formed^  into  a  vulgar  fraction^  by  dividing 
it  as  above  (12),  bi/  the  decimal  denominator  of  its  order , 
and  reducing. 

Examples  :  O.Qls"-  0.9l|  =  ^^==^^==.  1 1. 

0.07^5r=  0.7944=  Z^  =  _?Z5^=    7\ 
9  9         1000         11000       ^8' 

Addition,  subtraction,  multiplication,  and  division  of 

these  fractions,  w^hen  perfect  accuracy  is  desired,  are  best 

performed  by  changing  them  into  vulgar  fractions. 

Questions. — How  do  you  transform  a  vulgar  fraction  to  a  deci- 
mal one  ?  When  will  you  get  a  limited  number  of  decimals  ? 
How  many  decimals  will  you  have  in  that  case  ?  What  is  an 
infinite  fraction  ?  What  is  a  repeating  one  ?  Will  an  infinite 
division  always  produce  a  repeating  fraction  ?  How  do  you  indi- 
cate the  period  ?  If  you  do  not  reach  the  end  of  the  period,  how 
do  you  show  that  the  decimals  are  not  complete  ?  Does  the 
period  always  begin  after  the  decimal  point  ?  When  does  it  not  ? 
How  far  will  it  begin  from  it  ?  How  do  you  change  a  decimal 
fraction  into  a  vulgar  one  ?  How,  a  simple  repeating  one  ?  How, 
a  repeating  fraction  which  does  not  begin  at  the  decimal  point  ? 
How  many  decimals  will -/g- ;  -J^  '  8%'  "eVs^  &c.  give?  When 
w  ill  the  period  begin  in  the  decimals  of  -^j  5  -^j  5  -^-q  J    yV/  ; 


14?4«  LESSON  XL. 

EXERCISES. 

Transform  into  decimal  fractions  : 
1.      -i-V=.0625.  2.     ^VV  =  .0553  + 

3.     7oo^gg2^_^  ^^     111  =  1.03159557+ 

^*  "air 7  =  1.73930,  within  one  hundredth  of  thousandth. 

^*  TTT  •  •  to  the  tenth  place  of  decimals. 

*^*  TiT  •  •  to  within  one  millionth. 

8.  |-^    .  .  until  the  division  ends. 

^'  J^B  •  •  ^^til  the  period  is  formed. 

10.  tJj  .  . 

11.  ■g-V  =  0.'012345679. 

12.  Change  0.625;  0.550 ;  0.7250  j  0.21,  into  the  simplest  equi- 

valent vulgar  fractions. 

13.  What  fractions  are_equal  to  the  periods  0.162  ;  0.18  j   0.72; 

0.837   ^^27;  .02439;  0.0307692;  .03205128? 


LESSON  XL. 

REDUCTION  OF  FRACTIONS  TO  THE  SAME  DENO- 
MINATOR. 

1.  This  transformation  has  for  its  objiect  to  express  se- 
veral fractions,  having  different  denominators,  in  units 
of  the  same  kind.  It  is  a  necessary  operation,  prepara- 
tory to  the  addition  and  subtraction  of  fractions. 

For  instance,  we  can  add  ^,  ^,  and  j-,  because  they  are 
1,  2,  and  3  small  relative  units  of  the  same  kind,  namely, 
1  seventh,  2  sevenths,  3  sevenths,  whose  relation  to  a 
principal  unit  is  expressed  by  the  denominator,  7 ;  and 
we  get  -f +f +|-  =  -f-,  by  merely  adding  the  numerators; 
f  more  would  be  f,  and  so  on. 

2.  But,  if  we  had  to  add  ^  and  ^,  the  case  would  be 
different. 

The  units  being  dissimilar,  no  single  amount  could  be 
formed  of  them  (it  is  neither  2  halves  nor  2  thirds) ;  and 
some  transformation  is  necessary  to  reduce  them  to  ex- 
pressions of  similar  units. 


LEAST    COMMON   MULTIPLE.  14«5 

If,  for  example,  they  were  parts  of  feet : 


|-  might  be  called        .... 

6  inches, 

i       "           "               .... 

4     " 

And  the  sum  of  the  J  and  the  ^  of  a  foot } 

might   be  expressed  under  the  name   of  >  10  inches. 
inches,  as  .....) 

3.  With  abstract  numbers,  however,  the  relation  of 
units  being  altogether  numerical,  the  fractions  must  be 
transformed  into  expressions  numerically  equal. 

This  is  readily  accomplished  by  choosing,  for  a  com- 
mon denominator,  any  one  of  the  common  multiples  of 
the  given  denominators ;  such,  as  in  this  case,  as  6, 12, 18, 
24,  &c.,  either  of  which  may  be  assumed  for  the  com- 
mon denominator.  The  transformation  of  the  fractions 
will  then  be  made  (as  in  Lesson  XXX.,  6)  by  multiplying 
both  terms  of  each  by  the  factor  ivhich  will  change  its 
own  denominator  into  the  common  one. 

This  factor  is  evidently  the  quotient  of  the  new  by  the 
old  denominator. 

Thus,  according  to  the  object  in  view, 
5  and  -j  may  be  transformed  respectively  into 
i    and    I-  1  r  I 

or  j%  and  ^\  l"^^  ''"Zf^"\  fl     ^^^^^  "^^  ^^^  equivalent. 

oryVandy%j  [ff- 

&c.  &c.  &c. 

LEAST  COMMON  MULTIPLE. 

4.  Generally,  however,  the  simplest  form  is  preferred, 
and  the  least  multiple  chosen  for  the  new  denominator. 

In  the  preceding  example,  it  would  be  6. 

But  the  least  common  multiple  of  several  numbers  is 
not  always  so  evident,  and  a  systematic  method  must  be 
used  to  find  it. 

I  recommend  the  following,  which  depends  upon  this 
principle,  that  a  number  is  exactly  divisible  by  another 
when  it  contains  all  its  factors  (XXXVL,  1).     Therefore, 

5.  Decompose  the  largest  number  into  its  prime  fac* 
13  K 


14f6  LESSON   XL. 

tors;  then  examine  another  numher  ;  and ^  if  it  contain 
any  factors  beside  those  of  the  frst  numher ^  annex  them 
to  it  tvith  the  sign  of  multiplication  ;  do  the  same  things 
successively^  with  all  the  numbers. 

The  product  of  all  the  factors,  thus  connected^  will  be 
the  least  common  multiple  of  the  given  numbers. 

For,  it  is  evident  that  it  must  contain  every  number ; 
since  a  number  vi^hich  has  among  its  factors  those  of 
another,  contains  it :  216  contains  24,  because  it  has  3 
and  8  among  its  factors. 

It  is  the  least  common  multiple,  because  it  is  composed 
of  only  such  factors  as  are  indispensable  to  make  it  con- 
tain every  individual  number. 

6.  The  same  factor  is  frequently  found  several  times  in 
the  same  number,  and  should  consequently  appear,  in  the 
least  common  multiple,  as  many  times  (and  no  more)  as  it 
is  found  in  the  number  which  contains  it  the  greatest 
number  of  times  as  a  factor. 

In  other  words,  the  highest  power  (XXXIII.,  11)  of  any  one 
factor  must  always  appear  in  the  least  common  multiple. 

FIRST    EXAMPLE. 

Let  it  be  proposed  to  find  the  least  common  multiple 
between 

2,  3,  4,  6,  8,  9,  12,  18,  24,  36,  48. 

Take  48  as  the  largest,  and  decompose  it  into  its  sim- 
ple factors,  3x2x2x2x2;  or,  for  those  who  under- 
stand algebra, 3  X  2^ 

Now,  consider  the  second  number, 

36  =  3x3x2x2,  or  .  .  .  .  32x2". 
Since  it  contains   the   factor,   3,  once  more 

than  48,  we  must  annex  it  to  the  first  series 

of  factors,  which  increases  it  to 

3x3x2x2x2x2,     or  ...        3^x2*, 

which  now  contains  both  48  and  36. 

A  short  examination  shows  that  all  the  other  numbers 
are  contained  in  the  last  compound  multiple,  which  we 
conclude  to  be  the  smallest. 


LEAST    COMMON   MULTIPLE.  147 

The  multiplications  being  effected,  this  number  is  only 
144. 

That  it  is  the  least  common  multiple  sought  is»evident, 
since  it  contains  the  factors  of  all  the  numbers,  and  each 
factor  only  as  many  times  as  it  is  contained  in  that  num- 
ber which  contains  it  most  (that  is,  raised  to  its  highest 
power). 

SECOND  EXAMPLE. 

Let  us  now  take  the  numbers 

49;  21;  16;  75;  120;  840;  108;  112. 

The  largest,  decomposed  into  factors,  is 

7x3x2x2x2x5,     or      .         .         7x3x2^x5. 

Now,  120  is  evidently  part  of  it  already. 

112  being  7x2x  2x2x2,  or  7x2^  con- 
tains 2  once  more  as  a  factor.     There- 
fore, the  first  number  must  be  multi- 
plied by  it,  and  we  get 
7x3x2x2x2x2x5,     or         .         7x3x2'*x5. 

Considering,  now,  108,  which  is 

3x3x3x4,  or  3^X4,  we  have  to 
annex  to  the  common  multiple,  the  fac- 
tor 3,  twice,  by  which  it  is  increased  to 
7x3x3x3x2x2x2x2x5,     or     7x3^x2^x5. 

75  being  5  X  5  X  3,  we  have  for  it  an  addi- 
tional factor,  5,  which  raises  the  com- 
mon multiple  to 
7x3x3x3x2x2x2x2x5x5,  or  7x3'X  2^X52. 

To  contain  49,  it  requires  an  additional 
factor,  7,  and  becomes 
7x7x3x3x3x2x2x2x2x5x5, 
or 72x3^x2^x52,* 

which  completes  the  least  common  multiple,  since  16 

and  21  are  evidently  already  in  it.     The  product  being 

made,  gives  for  it       .    '     .         .         .         .         75,600. 

7.  In  practice,  so  minute  a  decomposition  is  not  neces- 
sary.    The  young  arithmetician  will  soon  learn  to  make 

*  The  simplicity  of  this  notation  recommends  it  to  those  who 
understand  it. 


148 


LESSON   XLI. 


most  of  the  analysis  mentally,  and  discover  readily,  not 
only  the  simple  prime  factors,  but  also  the  number  of 
times  th$y  are  to  appear  in  the  least  common  multiple 
(in  other  words,  the  power  to  which  each  is  to  be  raised). 
This  method  will  be  found  more  expeditious  and  improv- 
ing than  that  usually  adopted. 

Both  methods  require  a  ready  perception  of  the  divisi- 
bility of  numbers. 

Qzcestto7is. — What  is  the  object  of  the  reduction  of  fractions 
to  the  same  denominator?  Can  fractions,  having  the  same  deno- 
minator, be  added?  Why?  Can  fractions,  with  different  deno- 
minators, be  added?  What  must  be  done  first?  What  numbers 
will  answer  for  common  denominators  ?  Which  of  them  is  pre- 
ferable ?  How  is  the  least  common  multiple  found  ?  What  is 
necessary  in  order  that  a  number,  composed  of  factors,  may  con- 
tain another  ? 


EXERCISES. 

Find  the  least  common  multiple, 
1.     Of    3 J  6;  21. 


2. 

«      3;  5;  21;  35. 

3. 

f      8;  63;  560. 

Ans» 

5,040. 

4. 

'      8;  11;  13;  25;  43. 

Alls. 

1,229,800 

5.      < 

«    15;   11;  20;  21;  33;  55'y  120. 

Ans. 

1,320. 

6. 

«     4;  8;   12;  18;  24;  36. 

Ans, 

72. 

7. 

'«      9;  12;  18;  21;  25;  45;  54;  105; 
225.                 .... 

Ans, 

6,300. 

8. 

«    60;  28;  240;  225;  490;  720. 

Ans» 

176,400. 

9.      * 

'   20;    48;    280;    960;    1,800;    5,040; 
6,860.               .... 

A71S, 

4,939,200 

[0. 

<    11;  17;  19;  21;  7;  51;  187;  133; 
399 

Ans, 

74,613. 

LESSON  XLI. 

REDUCTION   OF  FRACTIONS  TO  THE  LEAST  COM- 
MON DENOMINATOR. 

1.  After  having  found  the  least  common  multiple  of 
the  denominators  of  several  fractions,  it  is  very  easy  to 


REDUCTION    OF    FRACTIONS.  149 

transform  them,  so  that  all  may  have  the  least  common 
multiple  for  their  denominator. 

This  is  a  transformation  by  augmentation,  which  it  has 
been  taught,  in  Lesson  XXX.,  is  made  by  multiplying 
both  terms  of  the  fraction  by  the  number  which  will  raise 
its  denominator  to  the  size  of  the  common  denominator ; 
iJds  number  is  the  quotient  of  the  least  common  multiple 
hy  the  denominator  of  the  fraction  (XXX.,  6). 

Let  us  take,  for  example,  | ;  |  ,*  ^  ;  /^  ;  \\ ;  g^-. 

The  least  common  multiple  is  readily  discovered  to  be 
96  ;  and  then  the  terms  of  the  various  fractions  must  be 
multiplied  each  as  follows  : 

2  X  32     3  X  24     7  X  12  _  5  X  4     _llj><_2       5X3 

3  X  32  '   4  X  24  '   8  X  12  '   24  X  4  '   48  X  2  '    32  X  3* 
The  operation  thus  indicated,  gives  at  last,  by  effecting 

the  multiplications, 

64.     72.     84.     20.     22.     15 
Ug"'    "9  6"  >    "gr  J    "SO"'    "9  6"  »    ■96"' 

for  the  fractions  reduced  to  their  least  common  denomi- 
nator. 

2.  The  decomposition  into  factors  of  the  least  common 
multiple,  facilitates  the  transformation  of  the  fractions, 
when  the  numbers  are  too  large  to  see,  at  once,  as  in  the 
preceding  example,  by  what  number  each  denominator  is 
to  be  multiplied.  For,  then  the  factors  of  each  denomi- 
nator being  known,  it  will  be  easily  discovered  by  what 
factors  it  must  be  multiplied,  to  be  raised  up  to  the  com- 
mon multiple,  and  it  will  be  sufficient  to 

3.  Multiply  both  terms  of  each  fraction  hy  all  thefac» 
tors  of  the  common  denominator,  which  are  not  in  its  own. 

Let  us  take,  for  example,  the  fractions 

17  .      U.-      17  3  .      319 
2B  )      2  4.  y     ^2 T  }      4T^' 

The  least  common  multiple  will  be  found  to  be 

72x3^x5^x2^; 

or    7x7x3x3x5x5x2x2x2  =  88,200; 
in  which  3,  5,  and  7  are  twice  factors,  and  2  three  times, 

13* 


150 


LESSON   XLI. 


Now,  if  we  compare  with  it  each  denominator,  we  find 
that 


28=2X2  X  7 
24  =2X2X2X3 
225  =5X5X3X3 
490=7  X7X5X2 


^or  ^ 


r22X7 
2^X3 
52x32 
72  X  5  X  2 


And  that  28^ 
24 
225 
490 


wants  the  factors 


'2X5X5X3X3X7  =  3,150. 
7  X7X3  X  1X5X5=3,675. 
7X7X2X2X2  =  392. 
3X3X5X2X2=  180. 


by  which  the  respective  fractions  must  be  multiplied, 
and  which  chano;es  them  into 


5  3.550. 


4  0,4?  5  . 


57,816  . 
"2  0^; 


33,420 
BH,200* 


5""g",2  0"^>     55,2irT5";    TS,2C 

Though  these  results  would  be  obtained  by  a  skilful 
calculator  without  so  much  detail,  I  would  even  recom- 
mend that  the  student  be  made  at  first  to  show  the  full 
arrangement  of  the  operations  j  placing  the  least  common 
divisor,  indicated  by  its  factors  as  above,  at  the  head  of 
his  work,  and  then 

17  X2X5X  5  X  3X3X7 
28  X  2X5X5  X  3  X  3  X  7  '' 
173X7X7X2X2X2 
225X7X7X2X2X2  5 

He  will  thus  understand  the  operation  better,  and  be 
made  sooner  sufficiently  familiar  with  it,  to  dispense  with 
the  dissection  of  the  denominators. 

Questions. — How  do  you  reduce  fractions  to  the  least  common 
denominator  ?  Repeat  the  rule  for  large  numbers.  By  what  do 
you  multiply  each  denominator?  Is  it  always  necessary  to 
decompose  the  fractions  into  factors  ? 


11X7X7X3X5X5 

24X7X7X3X5X.'5? 
319  X3X  3X5X2X2 

490X3X3X5X2X2' 


EXERCISES. 
Reduce  to  the  least  common  denominator : 


1  3.5.7. 

^'  4  ?    IT  ?    "S"  ? 

9  _2_  .     7  . 

*•  11  i    TJ 


.2.      4_ 

;  7     3  J     33* 


ADDITION    OF    FRACTIONS.  151 


2.4.       3    .      6     .       7    .       9 
y  ?    T  i    115  y    TT;    324  ?    40* 

3    .      4    .      2    .      6     .      7.11 
T3"?   TTJ    T5?   F^?    2^^   "JO' 


3. 
4. 

/r  7.       3.       1.13.      2.17.       43 

^'  15  i    TT?    5^7    24^   73^    5T  y    i2Ty 

ft  2.5.     11.     24.     11.        6_.     5^.     103 

^'  y";    TH^    2T?    2T7    45";    TX>5  ;    64?    '22T' 

„  1  1  .     _4    .       5    .     6  1  .     4  9  .    1 .     4  .     _7_ 

'•  20;     13?     2  6"?    "6T?    "62?     4?     5?     10* 

c  3    .    _6_.      4    .      2    .     12.      7..     JL3^  •    _1     . 

°'  22"?     121?    "SJ?    "99?     Tl?     iS?     195?     3  6T* 


LESSON  XLII. 
ADDITION  OF  FRACTIONS. 

1.  The  object  of  the  addition  of  fractions  is  to  find  a 
single  amount  equal  in  value  to  the  sum  of  several  frac-- 
tions. 

Questto??*  —  I",  :|,  |-,  ^  of  a  yard  have  lee7i  successively  cut 
from  a  yiece  of  cloth  :  how  many  yards  and  parts  of  a  yard  have 
been  take7i  out  of  the  piece  ? 

RULE. 

2.  I.  If  the  fractions  have  the  same  denominator,  add 
the  numerators  together,  and  place  their  sum  over  the 
common  denominator, 

n.  If  they  have  not  the  same  denominator,  begin  with 
reducing  them  to  the  least  common  denominator. 

3.  It  has  been  said,  in  Lesson  XL.,  that  we  may  add 
together 

1   4-  2     I     3 

because  they  are  units  of  the  same  magnitude :  and,  con- 
sequently, we  add 

1  seventh,    as  we  would  add     1  apple, 

2  sevenths,  "  2  apples, 

3  sevenths,  '^  3  apples. 

The  result  being  6  sevenths  in  one  case,  and  6  apples  in 
the  other. 

The  only  difference  being  that,  in  the  first  case,  the 


152  LESSON   XLII. 

name  of  the  units  has  reference  only  to  their  size,  as 
parts  of  the  same  thing,  or  unit  of  comparison,  and  is 
expressed  by  their  numerical  relation  to  that  thing. 

While,  in  the  second  case,  the  name  of  the  units  ex- 
presses merely  their  nature,  without  reference  to  their 
size. 

4.  But  if,  instead  of  units  of  equal  magnitude,  it  were 
proposed  to  add 

2.  4-    "^    4-5     3.7 

the  numbers  to  be  added  being  composed  of  thirds^ 
fourths,  sixths,  eighths—all  units  of  different  sizes — 
could  not  be  added  into  one  single  number.  Therefore, 
a  transformation  must  precede  the  addition,  in  order  to 
give  to  the  units  of  each  fraction  the  same  numerical  de- 
nomination, 

5.  The  least  common  de- 
nominator beins:  .... 


Arrange  the  operation  as  fol- 
lows, and  then  form  each 
numerator  according  to  Les- 
son XXX.,  by  multiply- 
ing the  old  one  by  the 
number  which  would  make 
its  denominator  24<. 


OPERATION. 

3X2X4=24 

2 
T 
3 
4 
5 
6" 

.  2X2X4=  16 
.  3  X3X2=18 
.       5X4     =20 

7X3     =21 

"75124 


Ssx  —  3s 


The  new  numerators  being  written  under  each  other, 
may  now  be  added.  Their  sum,  in  the  example,  is  75 ; 
and  the  result  might  be  written  |-|-. 

But,  generally,  it  is  changed  by  division  into  a  mixed 
number ;  and  the  fractional  part  reduced,  when  practica- 
ble, to  its  simplest  expression,  as  we  have  reduced  here 
li  to  3f  _ 

After  the  pupil  has  gained  some  experience,  he  may 
omit  the  intermediate  formation  by  factors  of  the  new 
numerators ;  but,  in  the  beginning,  the  complete  arrange- 
ment should  be  insisted  on. 


ADDITION   OF   FRACTIONS. 


153 


REMARK. 

6.  In  reducing  the  sum  of  the  fractions,  the  student  must 
observe  that  he  is  to  try,  in  the  numerator,  such  factors 
only  as  he  knows,  by  its  formation,  to  be  contained  in  the 
denominator. 

Thus,  in  the  preceding  example,  if  he  wished  to  re- 
duce If,  he  would  know  that  3  is  the  only  factor  to  be 
tried.     This  is  important  in  large  numbers. 

7.  Sometimes  the  fractions  are  not  alone,  and  form  part 
of  mixed  numbers,  to  be  added.     For  example: 


QUESTION. 


A  man  has  sold  the  following  number  of  yards  of 
cloth:  3i;  |:  5|-;  9|j  6i.  How  much  has  been  taken 
out  of  the  piece  1 

OPERATION. 


Least  common  denominator  2X2X3X5=  60 


Arrange  the  numbers  in 


si 

2 
q3 


1  X  15  =  15 

2  X  20  =  40 
5  X  10=50 

3  X  12  =  36 
1  X  30  =  30 


25^5 


171 


60 


qS  1   - 


this  way 

Add  the  fractions  as  di- 
rected above  ,*  their  sum 
is  here  yj  ;  take  out  the 
integral  part,  2,  and  carry 
it,  to  be  added  with  the 
other  whole  numbers,  and 
set  down  the  sum,  25,  to 
which  join  the  fractional 
part,  ij.  Thus,  the  number  of  yards  sold  is  ascertained 
to  be  2m. 

To  add  mixed  numbers,  add  the  fractions  first ;  re- 
duce  the  sum  ;  set  down  the  fractional  part,  a7id  carry 
the  whole  number  that  may  residt  from  their  addition  to 
the  column  of  whole  numbers. 

Some  direct  to  reduce  the  mixed  numbers  to  improper 
fractions ;  but  this  complicates  the  operation,  and  it  is 
always  absurd  to  form  large  numbers,  to  reduce  them 
afterwards. 


154  LESSON   XLII. 

SUBTRACTION  OF  FRACTIONS. 

8.  The  object  of  the  subtraction  of  fractions  is  to  find  a 
single  expression  for  the  difference  of  two  fractions. 

What  has  been  said,  in  regard  to  the  addition  of  frac- 
tions, applies  to  subtraction,  and  the  rule  is  similar. 

I.  If  the  fractions  have  the  same  denominator^  subtract 
the  less  from  the  greater  numerator,  and  set  the  differ- 
ence over  the  common  denominator, 

II.  If  they  have  not  the  same  denominator^  begin  with 
reducing  them  to  the  least  common  denominator. 

rp,         6  2  __  6-2  __  4. .   2  sevenths  taken  from 

>  T       T  7  7  7   6  sevenths^  leaves  4  sevenths  ; 

and  J  —  f  =  It"!!  ^^  /t>  by  reducing  the  two  given  frac- 
tions to  the  same  denominator,  24,  and  then  subtracting  the 
numerator,  16,  from  21. 

9.  When  tlie  fractions  are  joined  to  whole  numbers^  sub- 
tract the  fraction  first  and  then  the  whole  number, 

EXAMPLE. 


From      .    •    • 

.     13| 

Fraction  transformed 

f 

take    .     .     . 
The  result  is  .     . 

3-g; 

Difference      .     ,     .     . 

In  the  subtraction  of  mixed  numbers,  if  the  fractional 
part  of  the  number  to  be  subtracted  is  larger  than  the 
other,  the  operation  must  be  performed  by  means  of  an 
auxiliary  unit,  in  a  manner  similar  to  what  was  taught  in 
simple  subtraction. 


Let  us  suppose  that,  from  5| 

1  carried, 

we  have  to  subtract      3| 


OPERATION. 

J_      1  J.  _8_  —  1?  I   _§  —SO 
12      ■"  T^  1  2  —  12^12  ~'  72 

9  9 

12 T^ 


We  reduce  the  fractions  to  a  common  denominator ;  and, 
because  the  lower  one,  -^^,  is  larger  than  the  upper  one,  -^^^ 
we  add,  mentally,  to  the  fraction  of  the  upper  number,  a 


SUBTRACTION    OF    FRACTIONS.  155 

whole  unit,  which,  converted  into  a  fraction,  is  -}-|.  This, 
with  j\,  makes  f  §  ;  from  which  y^^  may  now  be  sub- 
tracted, leaving  ||. 

But,  because  we  have  thus  increased  the  upper  number 
by  1,  it  is  necessary  to  increase  also  the  lower  number 
by  the  same  amount,  in  order  that  the  difference  may  not 
be  altered.  This  is  done  by  carrying  one  to  the  3  units; 
which  makes  4  to  be  subtracted  from  5,  and  the  final 
remainder  is  IJ-J. 

It  is  not  necessary,  however,  to  add  -yl  to  j^.  In  practice, 
especially  when  the  fractions  are  large,  it  will  l^e  found  more 
simple  to  subtract  the  lower  fraction,  j^j  ^^om  the  transformed 
unit,  }  I,  and  then  add  the  remainder,  jo,  to  the  upper  fraction, 

8 

T2- 

10,  Here,  as  in  addition,  it  is  wrong  to  reduce  to  improper 
fractions  and  complicate  numbers,  which  are  to  be  simpli- 
fied afterwards. 

When  several  fractions  have  to  be  subtracted^  add 
them  together  fir  St  ^  and  subtract  their  sum, 

EXAMPLE. 

6l3_l  2         20     118 3 16.  —    3.8  -19  —    19 

6      '     4  8  3*       —  "2  4     »^  '2  4  24  24  54  24* 

Quesflo7is. — What  is  the  object  of  addition  of  fractions?  Of 
subtraction  ?  Repeat  the  rule  of  addition  of  fractions.  Why  do 
you  add  the  numerators  ?  Why  not  the  denominators  ?  If  the 
fractions  have  the  same  denominator,  what  is  the  nature  of  their 
units  ?  How  can  you  add  ^  and  5  ?  How  should  mixed  numbers 
be  added  ?  Give  the  rule  for  subtraction.  How  do  you  subtract 
fractions  having  the  same  denominator?  Having  different  deno- 
minators ?  How  do  you  subtract  mixed  numbers  ?  How,  when  the 
fraction  of  the  number  to  be  subtracted  is  the  larger  ?  Is  it  proper 
to  put  mixed  numbers  under  fractional  form,  either  for  addition 
or  subtraction?  How  do  you  subtract  several  fractions  from 
others? 

EXERCISES. 

1       2l34_5_L7j_23  —  Ji  1  7 
2.    H   +  ^3"  +  jio   +   604Xr  =  h 

3.  ?+l  +  f  +  |  +  |  =  2Ai. 


156  LESSON   XLIII. 

5.  9|  +  7|  +  5^  +  ll|.  =  35|. 

6.  7«+ll|  +  25i  +  9i  =  54}^-a. 

rj       19   13  —    63  cl6   8     8 

9.  14f~Mf=  10.  8f-4i=43^. 

11.  5|-2|  =  2J.  12.  3i-2i|=8|. 

13.  15 -f  = 

-lA  14-14-6  4    —  113 

1*r  1   JL-i-1   I4-I   23 

1«  5l3      2       —   10 

17.  l-]-|-i-TV-7\=^V 

18.  14TV-|-f-|-i  =  12,V 

19.  27J--3J--4|-7<-J=10|g. 

20.  234|  -  51  -  3>-  -  4f  -  117|  -  10  ji  =  92iJ. 

21.  Which  of  the  fractions  33-  and  y,  is  the  largest;  and 
what  is  their  difference  ?  ^,^5.  .3  __  _8_.  ==  1, 

22.  A  merchant  sells,  out  of  a  piece  of  cloth  measuring  SOj 
yards,  at  different  times,  7| ;  9f,  and  11 /^^  l^ow  much  must 
remain  ?  ^^^5.  23^4  yards, 

23.  There  is  a  pole,  |  of  it  in  the  mud,  |-  in  the  water :  how 
much  of  it  is  above  the  water  ? 

24.  Out  of  a  firkin  of  butter,  weighing  86;|-  pounds,  there  was 
sold  2^;  4J;  555  9f'  and  16  pounds.  The  balance  is  weighed, 
and  found  to  be  only  37-j  pounds  :  how  many  have  been  stolen  ? 

A71S,  10|. 

LESSON  XLIII. 

MULTIPLICATION  OF  FRACTIONS. 

1.  This  operation  is  an  immediate  consequence  of  the 
Twenty-ninth  Lesson. 

CASE   I. 

To  multiply  a  fraction  by  a  whole  number. 

RULE. 

Multiply  the  numerator  or  divide  the  denominator* 


MULTIPLICATION   OF   FRACTIONS.  157 

Question. — One  yard  costs  |-  of  a  dollar:  how  much  will 
5  yards  cost  ? 

They  will  cost  J  X  5  ==  ^-^  =  %'  =  4|. 

Second  example  :  ji  X  9  =  || ;  and,  by  reduction, 
would  be  y  =  ^2- 

Which  may  be  obtained  at  once  by  cancelling  9  with  18; 
or,  in  other  words,  applying  the  second  part  of  the  rule. 

This  has  been  explained  in  Lesson  XXX.,  9. 

CASE  II. 

2.  To  multiply  a  whole  number  by  a  fraction,  the  rule 
is  the  same  as  the  preceding ;  which  may  be  understood 
by  considering  that  -J  X  5,  is  the  same  thing  as  5  X  |-. 

3.  However,  as  the  principle  that  we  may  transpose 
the  multiplier,  has  been  proved  only  for  whole  numbers, 
it  might  not  be  admitted  for  fractions,  without  some 
demonstration. 

Let  us  multiply,  first,  5  by  7  whole  units ;  the  pro- 
duct is  either     .     .     5x7     or     7x5. 

But  this  product  is  too  large,  since  we  have  made  use  of 
a  multiplier  whose  units  should  be  divided  by  8. 

Now,  we  know  that  the  result  is  the  same,  whether 
we  divide  the  multiplier  or  the  product  (XXIIL,  3).  We 
do  the  latter ;  and,  dividing  by  8,  get  either 

-g —    or    — g—    for  the  correct  result. 

CASE    III. 

4.  To  multiply  two  fractions  together. 

Transform  mixed  numbers,  if  there  he  any,  into  equi~ 
valent  fractions  ;  then  multiply  the  numerators  together 
and  the  denominators  together. 

a        c         «  X  ^v 

(Algebraically,  3  X^=:^-^). 


DEMONSTRATION. 

Let  it  be  proposed  to  multiply  |  by  |. 
14 


158  LESSON    XLIII. 

If  we  had  to  multiply  1  by  the  whole  number,  5,  we 
would  multiply  the  numerator,  and  the  product  would 
be  (1)     3_X5 
4    • 

But,   in   multiplying   by  5    whole   units,   instead   of 

5  eighths^  we  make  the  multiplier  8  times  too  large. 

3X5 
The  product, ——,  is,  consequently,  also  8  times  too 

large,  and  must  now  be  divided  by  8 :  which  is  done  by 
multiplying  its  denominator,  4,  by  8  :  and  thus,  the  final 
result  is. 
The  product  of  the  numerators     ..3X5 

by  the  product  of  the  denominators   4X8        ^^* 
Which  proves  the  rule. 

This  is  the  general  case.  The  cases  of  multiplication 
of  a  fraction  and  a  whole  number  might  be  given  as  con- 
sequences :  since  the  product  by  the  whole  number  would 
require  no  rectification. 

5.  Though  made  part  of  the  rule,  the  transformation 
of  mixed  numbers  into  fractions  is  not  indispensable.  It 
is  frequently  preferable  to  multiply  by  parts. 

For  example,  to  multiply     .       .         .         .         16| 
•        by  .      ^.        .         18j 

Multiply,  first,  the  two  whole  numbers,  18  by  16, 

and  you  get 288 

Secondly,  f  by  18,  by  cancelling  9  out  of  18,  4 

Thirdly,  16  by  |,  by  cancelling  8  out  of  16,  6 

And,  finally,  |  by  f,  which,  by  cancelling  2  and 

o,  IS  •••••••  J  2 

Ansiver       •       298  j\j 

In  cases  like  this,  when  the  denominators  divide  the 
whole  numbers,  and  also  when  the  numbers  are  large,  this 
mode  of  multiplying  by  parts  is  preferable.  The  choice 
between  the  two  methods  must  be  regulated  by  the  judg- 
ment of  the  operator. 

6.  In  the  multiplication  of  vulgar  fractions,  we  ob- 


MULTIPLICATION    OF    FRACTIONS.  159 

serve,  as  in  decimals,  that  the  product  is  smaller  than  the 
factors.  We  can  explain  this  more  easily  still,  and  show 
that  the  rule  is  a  double  operation,  by  the  analysis  of  one 
of  the  questions  which  lead  to  multiplication  of  fractions. 

For  example : 

One  yard  of  stvff  costs  ^  of  a  dollar  :  what  will  f  of 
a  yard  cost  ?  If  two^thirds  of  a  yard  were  called  two 
feet^  there  is  no  one  who  would  think  of  multiplying  | 
by  2;  since  f  of  a  dollar  is  not  the  value  of  onefoot,hut 
of  one  yard;  and,  consequently,  not  the  true  multipli" 
cand  of  2  feet. 

But  the  value  of  one  foot  would  be  found  first,  by 
taking  one  third  of  that  of  a  yard;  and  then,  the  multi- 
plication being  properly  stated,  could  be  regularly  per- 
formed, and  the  product  would  be  found,  as  usual,  larger 
than  the  true  multiplicand. 

■g-  of  :|  of  a  dollar  is  \  :  this  is  the  value  of^  of  a  yard  ;  and 
twice  that  is  |  or  ^  of  a  dollar,  which  is  the  answer. 

Evidently,  whether  the  piece  of  stuff  be  called  |  of  a 
yard,  or  two  feet,  the  principle  is  the  same  ;  the  value  of 
^  must  be  obtained  first. 

7.  Therefore,  the  operation  is  compound ;  there  is  in 
it  both  a  division  of  the  multiplicand,  to  make  it  the 
value  of  one  unit  of  the  multiplier,  and  an  ordinary 
multiplication^  which  then  brings. a  result  larger  than  the 
true  multiplicand. 

In  the  present  example,  the  product,  ^  of  a  dollar,  is  larger 
than  the  altered  multiplicand,  i  of  a  dollar. 

In  practice,  both  operations  are  blended  together,  be- 
cause, dividing  the  product  gives  the  same  result  as  di- 
viding the  multiplicand ;  but  the  leading  operation  is  a 
multiplication. 

The  reading  of  the  numerical  statement  of  the  succes- 
sive operations  to  be  performed  shows  all  this  plainly.    It  is 

3X2      , 

4~v3~2'  Chy  cancelling  3,  and  then  2  out  of  4). 


160  LESSON   XLIII. 

Which  reads :  |  multiplied  by  2,  and  divided  by  3,  is 
equal  to  ^. 

FRACTIONS  OF  FRACTIONS. 
In  the  last  question,  we  have  taken  |  of  | ;  which  is 

Q 

I  divided  by  3 ;  equal  to   ^— — -    and  then  twice  that,  or 

f  X  f ,  is  evidently  taking  |  of  | ;  and,  when  so  ex- 
pressed, it  is  termed  a  fraction  of  fractio7i ;  and,  since 
the  same  operations  are  performed  as  in  the  multiplica- 
tion of  fractions,  it  follows  that  |  of  |  is  the  same  thing 
as  1  X  f . 

Hence,  fractions  of  fractions^  or  multiplication  of 
fractions^  are  identical  operations.  They  are  also  called 
compound  fractions  ;  three  names  for  the  same  thing. 

In  the  same  way  that  we  take  |  of  -|,  we  may  take  ^ 
of  the  result  j  that  is, 

and  then  another  part,  |,  of  the  result,  making  it  |  off  of 
I  of  f ,  &c.  All  of  which  will  be  another  mode  of  ex- 
pression for  multiplication  j  that  is, 

|of|of^of|  =  i|M|||^|. 

8.  Therefore :  A  compound  fraction  is  reduced  to  a 
simple  one  by  multiplying  the  numerators  together  and 
the  denominators  together. 

9.  In  practice,  indicate  all  the  products  and  divisions 
first,  as  in  the  above  example,  and  then  cancel  all  com- 
mon factors  between  the  compound  numerator  and  the 
denominator,  before  executing  the  multiplications. 

As  a  general  rule,  always  simplify  data  in  preference 
to  results  ;  it  will  shorten  the  operations. 

Here  you  see  that 

4  X2  ;  5  and  3  of  the  numerator  cancel  respectively 
8,       5  and  3  of  the  denominator,  and  thus  the  operation 
is  found  at  once  to  be  J,  without  going  through  any  multiplications. 


FRACTIONS  OF  FRACTIONS.  161 

N.  B. — Observe  that  cancelling  a  factor  means  dividing  by  it, 
leaving  in  its  place  the  quotient  1 ;  which  is  omitted  as  a  factor, 
because  it  does  not  alter  the  product  of  the  others  :  but  it  must, 
of  course,  be  set  down  as  a  numerator,  if  all  the  factors  happen 

4X2X5X3 
to  be  cancelled,  as  in  ^      o  w  o  ^^  4  =  i*     ^^t,  in  the  denomi- 
nator, it  would  be  useless,  since  division  by  1  gives  a  quotient 

2X3X4 

equal  to  the  dividend.    Thus,  — ^      ^      is  reduced  by  cancelling 

2  and  3,  to  4,  and  not  ^. 

It  is  not  uncommon  for  pupils.,  when  cancelling  a  factor,  to  say 
that  nothing  remains.  This  is  erroneous ;  they  should  say  that 
one  is  left,  though  not  set  down. 

Questions. — How  do  you  multiply  a  whole  number  and  a  frac- 
tion together ?  Why?  Show  that  the  result  is  the  same,  whether 
you  multiply  a  whole  number  by  a  fraction  or  the  fraction  by  the 
whole  number.  How  do  you  multiply  two  fractions  together? 
Demonstrate  the  rule.  How  do  you  multiply  mixed  numbers  ? 
Is  multiplication  of  fractions  a  simple  or  compound  rule  ?  Ex- 
plain it.  What  are  fractions  of  fractions?  How  are  they  reduced 
to  a  single  one?  What  should  be  done  before  performing  the 
multiplications  ?  If  you  strike  off  a  factor,  what  remains  in  its 
place  ?  Why  is  it  not  set  down  ?  What  is  f  of  | ;  |  of  4 ;  y 
of  5;    JJ  of  if;  f  of  \\l  &c. 

EXERCISES. 

1.  Multiply  9  by  |.  5.  Multiply  7  by  Sf. 

2.  Multiply  I  by   6.  6.    6|X7|  =  52^g. 

3.  Multiply  f  by  |.  7.   f  X  |  X  f  =  Jf 

4.  Multiply  4i  by  6.  8.   f  X  ^^^  x  5X  |  X|  =  4j. 

9.  f  of  4  multiplied  by  f  of  3f  =  ||. 

10.  5  X  I X  I  of  y  multiplied  by  4^-  =  S^-^. 

11.  I  of  I  of  I  of  f  of  12  =  3iV 

12.  4Sf  Xl3|•=:672xV 
13.   What  will  13^  gallons  of  wine  cost,  at  4 J  dollars  per  gal- 
lon ?  Ans»  64-g-  dollar s» 

14.  A  man  owns  y^y  of  a  cargo ;  \  of  which  are  lost :  what  is 
his  share  of  the  loss?  Ans,  -jq-. 

14*  L 


162 


LESSON    XLIV. 


15.  A  man  owns  2^5  of  a  capital,  and  sells  -j^o  of  this  part : 
what  part  of  the. capital  does  he  sell?  ^        A72S.  -^^j. 

16.  One  pound  of  sugar  costs  -]-|  of  a  dollar :  what  will  17| 
pounds  cost  ?  A7is.  IGy^g  of  a  dollar. 

17.  One  yard  costs  6fjj  of  a  dollar  :  what  will  f  cost  ? 

Ans.  4?  dollars* 

18.  A  man  failing  in  trade,  can  pay  only  |  of  a  dollar  on  each 
dollar :  how  much  will  he  pay  on  122"  dollars  ?      Ans.  5  dollars* 


LESSON  XLIV. 
DIVISION  OF  FRACTIONS. 

CASE   I. 

1.  To  divide  a  fraction  by  a  tvhole  number^  divide  the 
nrimerator,  or  multiply  the  denominator  (with  algebraical 
symbols  f  •  ^  =  ^)- 

This  is  a  repetition  of  what  was  said  in  the  First  Part, 
Lesson  XXIX. 

1st  Question, — 3  yards  have  cost  f  of  a  dollar,  how  much  is  it 
a  yard  ?    Here  you  divide  the  numerator.         A71S.  1  of  a  dollar, 

2d  Question. — 5  yards  have  cost  -I  of  a  dollar,  what  is  it  a 
yard  ?  Here  you  cannot  divide  the  numerator,  and  must  multiply 
the  denominator.  A?is.    3    of  a  dollar* 

CASE  II. 

2.  To  divide  a  whole  number  by  a  fraction^  multiply 
the  number  by  the  denominator  and  divide  by  the  nume^ 
rator  (with  algebraical  symbols  a  :  |  =  f ). 

DEMONSTRATION. 

Let  it  be  proposed  to  divide  12  by  •^;  that  is,  to  divide 
12  by  7  units  of  a  size  referred  by  the  denominator  9  to 
another  known  unit. 

If  we  had  only  to  divide  12  by  7  the  result  would  be 


DIVISION    OF    FRACTIONS.  X^ 

but  the  quotient  is  not  7  whole  units,  but  a  number  of 
units  9  times  smaller.  The  divisor  we  have  used  is 
consequently  9  times  too  large;  and,  consequently, 
(XXIV.,  3)  the  quotient  would  be  9  times  too  small  with- 
out a  correction  :  in  order  to  obtain  the  right  result, 
therefore,  we  must  now  make  this  quotient  9  times 
larger  by  multiplying  the  numerator,  which  gives  for  the 
final  result  ^ 

12X9        108        ^^3  * 

—7 —  =  -7-  =  I'^y 

If  the  question  had  been  to  divide  12  by  -f ,  the  numerical  state- 

12  X  5 

ment,  either  written  or  mentally  arranged,  would  be  — -r—  •    ^^ 

this  case  4  can  be  cancelled,  and  the  result  would  be  3X5  =  15. 

3.  The  operation  evidently  amounts  to  multiplying  the 
whole  number  by  the  fraction  inverted  (algebraically 
«:-*  =  «  X  4). 

4.  In  both  cases,  the  quotient  is  larger  than  the  divi- 
dend, because  a  thing  smaller  than  one  must  be  contained 
in  any  number  oftener  than  one  itself. 

5.  To  divide  one  fraction  by  another,  multiply  the 
dividend  fraction   by  the  divisor  inverted  (algebraically 

a  ^  c  <*   V    ^\ 

DEMONSTRATION. 

Let  it  be  proposed  to  divide,  for  example,  f  by  -^j. 

We  might  reason  as  in  the  preceding  case,  but  we  will 
vary  the  demonstration,  and  show  another  way  to  arrive 
at  the  result. 

Make  the  divisor  a  whole  number  by  striking  off  its 
denominator  11.  This  amounts  to  multiplying  the  divi- 
sor by  11  (XXIX.,  5).^ 

Having  thus  multiplied  the  divisor,  we  must  also  mul- 
tiply the  dividend  by  11,  in  order  to  get  the  same  quo- 
tient (XXIV.,  4),  which  gives  us  the  new  dividend 

3  X  11 
5 

to  be  divided  by  8 ;  and  the  final  division  is  effected  as 


164  LESSON   XLIV. 

above  by  multiplying  the  denominator  by  8,  so  that  the 
result  is 

3  X  11      3       n 

5  X   8  ^'^  5  ^    8  » 

the  latter  being  the  divisor  inverted. 

6.  Before  effecting  the  operations,  be  careful  to  cancel 
the  common  factors : 

16 
If  we  had  If  :  j-j ;  or,  with  the  other  sign  of  division,  ^ 
the  indicated  operations  would  be  -j  5 

1AX^5  =  (by  cancelling)!. 

Observe  that  cancelling,  in  the  division  of  two  frac' 
iions  reduced  to  their  simplest  terms,  takes  place  between 
numerator  and  numerator,  and  between  denominator  and 
denominator. 

In  the  preceding  example,  8  is  cancelled  out  of  16,  and  5  out 
of  both  15  and  25. 

7.  Had  the  divisor  been  |  instead  of  -f^,  the  result 
would  have  been 

16    V    5   —  4 

Which  shows  that  the  division  of  fractions  may  he  made 
by  dividing  numerator  by  numerator,  and  also  denomi- 
nator by  denominator,  when  they  will  divide. 

Hence,  if  the  denominators  are  the  same,  we  have 
only  to  divide  the  numerators  j  since  equal  numbers 
cancel. 

8.  If  either  the  dividend  or  the  divisor,  or  both,  are 
mixed  numbers,  transform  them,  first  into  fractional  ex- 
pressions, and  then  divide  as  above. 

If  we  had  to  divide  12|  by  6f ,  the  successive  operation  would  be 
12j-  :  6|  =  V  :  \";  -'  |-==?>S-0=  ^'^  =  '^- 

9.  It  will  be  observed  that  the  four  fundamental  rules 
m  fractions  require  no  new  principles,  and  are  mere 


DIVISION    OF    FRACTIONS.  1^ 

combinations  of  the  simple  rules  of  the  First  Part.    They 
are,  in  fact,  compound  operations. 

In  division,  as  well  as  in  multiplication  by  a  fraction, 
we  remark  that  the  operation  requires  both  a  multiplica- 
tion and  a  division.     For,  the  division  of  12  by  |-,  for 

12  ^  9 
example,  is  indicated  thus :   ■ — - — -^  which  reads,  12  twwZ- 

tiplied  by  9,  and  divided  by  7. 

10.  Here  we  observe  a  similarity  of  features,  which 
makes  it  sometimes  difficult  for  learners  to  distinguish,  in 
combinations  of  fractions,  whether  they  have  to  multiply 
or  to  divide.  They  will  remove  all  doubt  by  reading  the 
question  with  whole  numbers.  The  operation  in  fractions 
will  be  analogous  to  that  with  integers  j  which  will  be 
recognised  by  considering  that : 

1st.  If  one  being  given,  many  are  to  be  found,  it  is 
7nultipUcation, 

2d.  If  a  whole  being  given,  a  part  is  either  given  or 
required  5  in  other  words,  if  a  product  and  one  factor 
being  given,  the  other  is  required,  it  is  division. 

For  instance : 

1st  Question. — One  yard  costs  -f  of  a  dollar;  how  much  will  § 
of  a  yard  cost  ?  If  we  read  without  the  denominators,  as  fol- 
lows :  One  yard  costs  3  dollars  ;  how  much  will  2  yards  cost?  we 
see  at  once  that  it  is  a  multiplication. 

2d  Question. — How  much  stuff  can  be  bought  for  I-  of  ^  dollar, 
at  "I  of  a  dollar  a  yard  ?  Read  without  the  denominators,  and 
you  easily  perceive  that  the  amount  to  be  expended  is  a  dividend, 
and  the  cost  per  yard  its  divisor.  Consequently,  the  analogy 
shows  that  the  operation  is  a  division. 

3d  Qiiestion. —  y  of  the  income  of  a  man  amounts  to  400  dol- 
lars :  what  is  his  income  ?  Read,  without  the  denominator, 
4  times  the  income  of  a  moAi  is  400  dollars :  what  is  the  income  ? 
Evidently,  you  must  divide  400  by  4;  consequently,  you  must 
also  divide  400  dollars  by  ±, 

Questions. — How  is  a  fraction  divided  by  a  whole  number? 
Why  ?  A  whole  number  by  a  fraction?  Why?  A  fraction  by 
a  fraction?  Why?  What  must  be  done  before  effecting  the 
operations  ?  When  the  numerator  of  the  dividend  is  a  multiple 
of  that  of  the  divisor,  what  is  done  ?     When  the  denominator  of 


166 


LESSON   XLIV. 


the  dividend  is  a  multiple  of  that  of  the  divisor,  what  is  done? 
Hovr  do  you  divide  mixed  numbers  ?  Is  division  of  fractions  a 
simple  or  compound  rule?  How^  will  you  ascertain  whether  the 
question  is  a  multiplication  or  a  division?  Give  examples.  la 
J  of  36  a  multiplication  or  division  ? 

EXERCISES. 
All  the  results  must  be  reduced  to  their  simplest  expression. 


1.  Divide  I    by  |. 

2.  if  by  f . 

3.  2%  by  if. 


4. 

Vbyf 

5. 

/«by|. 

6. 

fof  ibyf. 

7. 

7}by9|=|f. 

8. 

fof|--fof7|- 

=  T^T 

9. 

S=ii- 

10.  ?f  = 

11.  -f  = 

10 

13.^= 
3cr 


100 

4^ 


13. 


=  20|t. 


15.   5,205}    :   I  of  91  =  7l| 


14.    6| 


2i  = 


16.  At  ;|  dollar  a  bushel,  how  many  bushels  can  be  had  for 
5  dollars  ?  A?is.  6^  bushels, 

17.  By  how  much  must  3 J  be  multiplied,  that  the  product 
may  be  291  ]  Ans.  9. 

18.  What  part  of  I  is  f? 

19.  What  part  of  7f  is  2J1  Ans.  4|f . 

20.  At  ^  of  a  dollar  per  bushel,  how  much  can  you  buy  for  § 
of  a  dollar  ? 

21.  If  a  man  can  build  5|  yards  in  a  day,  in  how  many  days 


will  he  build  18?  ? 


A71S.  3g  days* 


22.  If  it  take  f  of  a  bushel  to  sow  an  acre,  how  many  acres 
will  16  bushels  sow? 

23.  If  -J  of  a  gallon  cost  3|-  dollars,  how  much  is  it  a  gallon  ? 


MISCELLANEOUS  QUESTIONS  IN  FRACTIONS  AND  CANCEL- 
LING. 

1.  A  yard  costs  47 1 :  what  is  the  value  of  12 1  yards  ? 

Ans»  OlOjig  dollars. 


DIVISION    OF   FRACTIONS. 


167 


2.  A  person  has  bought  23^5_  yards,  for  745 is  dollars  :    how 
much  is  it  for  a  yard  ?  Ans.  $31.86  + 

In  the  following  exercises,  recollect  that  cancelling  may  take 
place  between  numerators  and  between  denominators  of  frac- 
tions, which  are  to  be  divided  by  each  other ;  and,  therefore, 
cancel  where  you  can  before  combining  the  numbers.     Example : 


i  \  £ 

7^^^21 


tX) 


=  3. 


6. 


y   X  3g  X  y^ 


TT^  H^ 


TT 

■f  X|X3 

ttX|X4 


i-^  =  JIl_ 

9  135* 

2 


=  tV2. 


5. 


V. 


2i  +  lJ^54 


3|-4 


4J-  —  24 

^ =  -^A. 

6,1  -24-        ^22 


10. 


11. 


1  4-  A  —  i. 

•'•12  3 

I  +  0.453 
0.46 

5|  +  7.35 


4:  7 


4.375 
4.37- 


_  1 
2 

1.75 


4J 
0.12 
12.    2.75 


.9I 

^2 


=  2.8869  + 
3.39354+ 
=  1.1657 


+  1.314=0.67881-1- 


13.  In  a  book  is  found  |  +  +  f  +  1  +  Ts  =  ^iVV- 

The  second  number  being  defaced,  what  should  it  be  ?      Ans,  -f-. 

:  |- :  what  is 


14.  There   is  also   found  -j^q-  of  ~ 
the  missing  denominator  ? 

15.  Likewise,  9f  divided  by  ^  of  . 
is  omitted  ? 


of  I  of  I 


.  .  .  =  2JI :  what  number 
Ans,  7. 

16.  Of  what  number  is  176  the  \\  part  ? 

17.  ^  owns  f  of  4-  of  a  ship,  and  j5  |  of  |  :  which  is  the 
greater  share,  and  by  how  much  ?  Ans,  A  has  y  tnore, 

18.  A  farmer  had  his  sheep  in  five  fields ;  in  the  first,  he  had 
\  of  them ;  in  the  second,  \ ;  in  the  third,  \ ;  in  the  fourth,  y^ ; 
and  in  the  fifth,  450 :  how  many  had  he  in  all  ?     Ans,  1200  sheejp. 


168  LESSON  XLV. 

19.  f  of  the  income  of  a  gentleman  are  $4,593;  "what  is  the 
\7hole  income  ?  Ans. 

20.  A  merchant  loses  §  of  his  private  income,  "which  amounts 
to  £283^,  but  he  receives  from  his  business  £568^,  "which  is  a  profit 
of  §  on  his  capital:  how  much  does  he  get  in  all ;  and  "what  is  his 
capital  in  trade  ?  Isi  Ans.  £770 J.     2d  Aiis.  £1420f . 

21.  A  grocer  sold  ^  of  a  gallon  of  "wine  for  -fV  ^^  ^  dollar: 
"what  "Was  it  a  gallon?     Is  this  multiplication  or  division ? 

A71&'.    2-|  dollars, 

22.  A  man  travels  4  miles  in  -g-  of  an  hour :  hov^  much  does  he 
travel  an  hour  ?  A?is.  4j  miles. 

23.  "I  of  a  pole  is  6|-  ft.  long :  hov^^  long  is  the  pole?     A7i.s. 

24.  A  pole  is  |  in  the  mud,  f  in  the  water,  and  4 1%  ft.  above 
the  "water  :  how  long  is  the  pole  ?  A7is»  12^^  ft. 

25.  A  person  having  spent  ^  of  his  money  at  one  time,  and  -g- 
at  another,  had  2G§  dollars  left :  how  much  had  he  ? 

Ans.  160  dollars. 

26.  135  J  |-  is  iV  of  what  number  ?  A7is.  24l|-f  • 

27.  A  man  owns  |  of  a  ship,  and  sells  -J  of  his  share  for  4000 
dollars:  what  is  the  value  of  the  ship?  Atis.  15,000  dollars. 


LESSON  XLV. 

(Which  may  be  omitted  "with  beginners.) 

SHORT  METHODS  IN  MULTIPLICATION. 

The  following  transformations  will  greatly  simplify- 
both  written  and  mental  operations. 

1.  To  multiply  by  5,  multiply  hy  10  and  divide  by  2. 
Because  5  =  ^^. 

Examples:     673X5  =  ?^=  336.5;  97.57X5  =.5Z|ll=: 


4S7.85;    4.17X0.05  =  tll><^  =  0.2085 


2 


In  this  and  following  examples,  it  will  be  remarked  that  the 


SHORT    METHODS    IN    MULTIPLICATION.  169 

multiplication  by  10;  100;  1000,  &C.5  requires  merely  the  dis- 
placing of  the  units'  point,  either  before  or  after  multiplying.  In 
this  consists  the  advantage  of  these  methods. 

2.  To  multiply  hy  25,  multiply  by  100  and  divide  by  4;. 

Because  25  =  ^f  ^. 

Examples  :    347X25  =  ^^1^  =  8675  ;  67.4X25  =  5Zi5  = 
4  4 

1685;   16.77X0.025  =  1^''77X0.100  _   4^925. 
4 

3.  To  multiply  by  50,  multiply  by  100  and  divide  by  2, 
This  is  nearly  the  same  thing  as  the  first  case. 

4.  To  multiply  by  75,  multiply  by  100  and  deduct 
one-fourth  of  the  product.     Because  75  =  100  —  J^|^. 

Examples:  35X75=  3500  — ?£^  ==  2,625;  0.29X0.75  =  .29 
—  0.0725  =.2175. 

5.  To  midtiply  by  125,  multiply  by  1000  and  divide 
by  8.     Because  125  =  ii>gO-\ 

Exampks :    38 X 125  =  ^?^-£^  =  4,750 ;     4.9  X 1-25  =i?  = 
8  8 

6.125. 

6.  To  multiply  by  12^,  multiply  by  100  and  divide 
by  8.  This  is  about  the  same  rule  as  the  preceding,  by 
writing  12^  decimally,  12.5. 

7.  To  multiply  by  37^,  multiply  by  100  and  take  f 
of  the  product.     Because  37^  ==  |  of  100. 

Example:     457  X  37 1  =   i?™?  =  17,137.5. 

8.  To  multiply  by  62j,  multiply  by  1000  and  divide 
hy  16.     Because  62j  =  ^^^. 

Examples:    129X62J  =   1?55H  =  8,062.5;  27.35X6.25  =ff 
16 

16 

15  - 


170  LESSON   XLV. 

9.   To  multiply  hy  87 1,  multiply  hy  100  and  talce  |  of 
the  product ;  or,  deduct  ^  of  the  same, 

13900X7   _  97300 
~8       '  8 


Example  :     139  X87 1-  =:    ^'^--^  ^  '   =  11:^  =  12,162.5;  or, 


13900—   H555  =  12,162.5. 
8 


10.  To  multiply  by  16|,  multiply  by  100  and  divide 
by  6.     Because  16f  =  JL^j.. 

Example:    36X161  =  ^^  =  600. 
^  6 

11.  To  multiply  by  33^,  multiply  by  100  and  divide 
by  3.     Because  33|  =  ^K 

Examples:  48X333-  =  1,600;    7.67X0.33^  =  2.556. 

12.  To  midtiply  by  66|,  multiply  by  100  and  talce  | 
of  the  product. 

Example,  78X66|  =  i^  =  5,200;  or,  =  2,600X2  = 
5,200. 

13.   To  multiply  by  15,  "^  you  may    [    30,    and  talce  one 
35,  I  midtiply     \     70,    half   of  the 

45,   [  respective-  \      QQ       product, 

ovbb,]^y^y      [no. 

Or  else^  you  may  take  one-half  of  the  multiplicand 
before  midtiplying. 

Examples:     38X15=19X30  =  570;  475X35  =  ^'^^  '    = 

S?H  _  16,625. 
2       ~" 

76X45  =  38X90=3,420;     57X55=51^111?  =  .5?2^  = 

2  2 

3,135. 

You  may  also  multiply  by  15,  by  adding  one-half  of  the  num- 
ber to  it  and  multiplying  by  10  (or  inversely). 

Thus:  86X15=  (86  +  43)  X10=  1,290;  79X15  =  790  +  395 
s=  1,185. 

14.  As  regards  multiplications  by  such  numbers  as  18, 


SHORT    METHODS   IN   MULTIPLICATION.  I7l 

24,  36,  48,  63,  &C.5  which  can  be  decomposed  into  fac- 
tors, I  refer  to  contraction  in  multiplication  (XXXI.,  4). 

15.  To  multiply  by  11,  you  may  add  each  figure  to 
that  on  its  right. 

To  avoid  mistakes  in  applying  this  rule,  beginners  will 
do  well  to  conceive  a  zero  to  be  placed  at  each  extremity 
of  the  number,  and  use  them  as  figures  of  the  number. 

Example:      789 XH*      Conceive  the  number  to   be  written 

07890,  and  then  say 9+0=9. 

8  +  9=17;  carry  1  and  set  down        ...         7. 
1  carried  +  7  +  8=  16;  carry  1  and  set  down        .     6. 

lcarried  +  0  +  7 =8. 

So  that  the  whole  product  is  8,679. 

The  reason  of  this  operation  will  be  easily  discovered 
by  performing  it  in  the  regular  way.  A  direct  multipli- 
cation by  11  will,  however,  be  nearly  as  expeditious. 

16.  To  multiply  by  any  number  between  12  and  20, 
you  may  multiply  successively  each  figure  of  the  multi- 
plicand by  the  last  of  the  multiplier^  and  add  to  each 
^product  the  figure  of  the  7nultiplicand  on  the  right  of 
that  you  multiply. 

Example:     423X17. 

7X3=  21,  carry  2  and  set  down 1. 

7X2  +  2  carried +  3=  19;  carry  1  and  set  down         .  .     9. 

7  X4  +  1  carried  +2  =  31;  carry  3  and  set  down              .  1. 

3  carried  +  4,  the  extreme  left-hand  figure             .         •  .7. 

So  that  the  product  is  7,191. 

In  the  case  of  the  multiplier,  19,  it  might  be  preferred 
to  multiply  by  20,  and  deduct  the  given  multiplicand. 

Thus:  365X19=365X20  —  365. 

17.  This  remark  applies  also  to  29,  39,  49,  &c.  When 
you  have  such  multipliers,  you  may  multiply  by  30,  40, 
50,  &C.5  and  subtract  one  multiplicand. 


OPERATION. 

914785 
246144 


5488710.  ..  product  by  6. 

21954840....  this  by.  .  4. 

181729040  and  then  by  6. 


225168839040 


172  LESSON   XLV. 

MULTIPLICATION  BY  MULTIPLES. 

18.  Suppose  we  had  to  multiply  914,785  by  246,144. 
It  is  indifferent  by  which 
figure  we  multiply  first 
(XIIL,  5).  Therefore,  re- 
marking that  24  is  a  mul- 
tiple of  6,  and  144  a 
multiple  of  24,  we  multi- 
ply first  by  6,  observing  to 
set  the  product  in  its  right 
place ;  then  we  multiply 
this  first  product  by  4, 
which  gives  that  by  24; 
which,  finally,  we  multiply  by  6,  to  get  that  by  144. 

If  the  last  part  had  been  145,  instead  of  144,  we 
might  still  have  proceeded  in  the  same  wa}'-,  and  added 
one  multiplicand;  if  150,  the  product  by  6  should  have 
been  added. 

This  example  will  suffice.  The  skilful  application  of 
the  method  will  depend  on  the  readiness  of  the  operator. 

SHORT  METHODS  IN  DIVISION. 

Shorter  methods  are  less  numerous,  and  also  less  useful, 
in  division  than  in  multiplication. 

19.  To  divide  by  5,  multiply  by  2  and  divide  by  10. 

Examples:    IZ^  =  Yl^^  =  35.2;      ^2^  =  ^^^'^X'^  = 
5  10  0.05  0.10 

7,594. 

20.  To  divide  by  25,  multiply  by  4  and  divide  by  100, 

Examples  :    H^  =  5?^><i  =  11.52  ;      .^   =  ^_:^XJ  = 
25  100  0.025  0.100 

254.4. 

21.  To  divide  by  125,  multiply  by  8  and  divide  bv 
1000. 

22.  «  by  12^,  multiply  by  8  and  divide  by 
100. 


SHORT   METHODS    IN    DIVISION.  173 

23.  To  divide  by  62.1,  multiply  by  16  and  divide  by 
1000. 

24.  "  by  16|,  multiply  by  6  and  divide  by 
100. 

25.  "        by  33j,  multiply  by  3  and  divide  by 

100. 

26.  "  by  37|,  multiply  by  8  a/i^Z  divide  by 
3  anci  100. 

27.  "  Z>?/  87|,  acZ^  one-seventh  and  divide  by 
100.    Because  87  ^  +  ?Zi  =  100. 

2  rj 

Ecca^nple:   ^  =  ^^±^  _  7.84. 
87J-  100       - 

28.  To  divide  by  66^,  add  one-half  and  divide  by 
100.     Because  66^  +  ^  :=  100. 

E:can.ple:   ^   =  ^±i^   =1.02. 
66|  100 

29.  To  divide  by  15,  35,  45  or  55,  double  the  divi" 
dend  and  divide  respectively  by  30,  70,  90,  110. 

30.  To  divide  by  75,  add  one-third  and  divide  by  100. 
Because  75  +  ^  =  100. 

„  ;       237         237  +  79         .>  i« 

Example:   =  ! .  =:  3. Id. 

^  75  100 

I  have  introduced  these  rules  here,  in  order  that  they 
may  be  used  in  subsequent  operations.  Algebra  will 
furnish  other  ready  ways  of  simplifying  operations. 

Qitestions. — What  simple  way  is  there  to  multiply  by  5 ;  25; 
50,  &c.  .  .  .  ?  How  do  you  multiply  when  multiples  are  disco> 
vered  in  the  multiplier  ?  What  short  method  is  there  to  divide 
by  5;  25,  &c.  .  .  .  ? 

EXERCISES. 

Exercises  on  the  preceding  methods  can  be  performed  with 
15* 


174^ 


LESSON   XLVI. 


advantage  only  under  the  eye  of  the  teacher,  who  can  readily 
supply  them.  The  pupil  may,  however,  perform  the  following 
by  himself: 

Multiply,  by  making  use  of  such  multiples  as  may  exist  in  the 
multiplier, 

1.  6,927  by  567  =  6,  942,566  by  64,816  = 

2.  79,088  by  18,729=  6.  6,720,701  by  3,361,216  = 

3.  687,846  by  168,287=  7.  265,603  by  525,625  = 

4.  963,792  by  1,296,133  =  8.  345,784  by  6,752,459  = 


LESSON  XLVL 
PROPORTIONS.* 
3  :  6  :  :  18  :  36     read    3  is  to  6  as  18  is  to  36. 
Syjnbols.—a  :  b  :  :  c    :  d        "        a     "     5     "  c      «      d. 

Questio7i. — If  3  yards  cost  6  dollars,  how  much  will  18  yards 
cost  ?  Ans,  36  dollars » 

1.  This  subject  is  introduced  here,  because  proportions 
^are  nothing  but  fraction?^  in  disguise;  and  another  con- 

^  I  •    fusion  of  terms  for  things  already  known,  serves  to  make 

I  !   a  separate  theory,  which  does  not  add  an  iota  to  the 

)  knowledge  of  arithmetic.    • 

'  The  doctrine  of  proportions  might  be  altogether  dis- 

.  pensed  with,  even  in  geometry,  where  they  are  most 

;  useful. 

2.  It  has  been  said  (XXX.,  5)  that  the  same  fraction 
might  be  transformed  in  an  infinite  number  of  ways; 
such  as 

1  .      9,      3  .      18.     _5_6_      &p    . 

2  7      4  ?      "6   7       36' 7      1125     ^"^^  7 

whicli  are  all  equal  in  value ;  that  is,^th£LJIumerator  and 
denominator  of  everyone  have  the  same  ratio  to  each 
jDth^rT    So  that,  lor  example^" 

*  Formerly  two  sorts  of  proportions  were  used  ;  arithmetical  and 

feometrical  proportions.  The  first  has  been  pretty  nearly  abandoned, 
lere  we  consider  only  geometrical  proportions,  which  will  proba- 
bly in  time  be  restored  to  the  doctrine  of  fractions,  to  which  they 
belong. 


PROPORTIONS.  179 

3  —    18. 
'6  —  Te  ) 

or,  the  ratio  of  3  to  6  is  the  same  as  that  of  18  to  36. 

3.  Let  it  be  remembered  that  the  ratio  of  two  quanti- 
ties is  the  quotient  of  the  second  by  the  first  (XXX.,  4) ; 
in  the  examples  just  preceding,  the  common  ratio  is  2. 

4.  Two  equal  ratios,  or  fractions,  written  with  the 
other  sign  of  division  ;  thus, 

3  :  6  =  18  :  36,     or     3  :  6  :  :  18  :  36, 

are  called  a  proportion ;  which  is  read  as  above,  Sis  to  6 
as  IS  is  to  36  j  or,  as  3  is  to  6,  so  is  18  to  36.     Hence, 

5.  A  PROPORTION  is  the  expression  of  the  equality  of 
two  ratios. 

Would  not  I  =  J-f,  be  just  as  plain  and  more  in  accord- 
ance with  preceding  rules ;  and  could  it  not  be  read  ex- 
actly in  the  same  way?  I  recommend,  indeed,  to  pupils 
to  accustom  themselves  to  write  proportions  so. 

6.  This  is  not  all ;  new  names  are  also  given  to  the 
numbers  which  compose  the  proportion. 

The  1st  number  is  called      .  1st  antecedent. 

the  2d         "  "  .  its  consequent. 

the  3d         "  "  .  the  2d  antecedent. 

and  the  4th        ''  "  .  the  2d  consequent. 

It  is -therefore  said  that  the  1st  antecedent  is  to  its  con- 
sequent as  the  2d  antecedent  is  to  the  2d  consequent. 

The  four  numbers  are  the  terms  of  the  proportion ;  the 
first  and  last  are  the  extremes;  and  the  two  middle  ones 
the  means. 
So  that, 
the  dividend,  ^  i  2     (  the  numerator,     1  |     f  the  antecedent ^ 
the  divisor,     \^%f<,^  the  denominator,  Y^t\  the  consequent ^ 
and  the  quotieiit,  J  f  J  |  [and  the  fraction,)  ^1  [and  the  ratio. 

7.  Besides  their  fundamental  property,  that  the  two 
ratios  are  equal,  proportions  have  another  equally  essen- 
tial property,  viz  : 

The  product  of  the  extremes  is  equal  to  the  product  of 
the  means. 


176  LESSON   XLVI. 

Thus, 

in  the  proportion,      3  :  6  :  :  IS  :  36,  we  have  3  X  36  =  6  X  18. 
and  with  symbols,  ina  :  b  :  :   c  :   d,         "        a  X  d  =  b  X  c, 

DEMONSTRATION. 

This  property  can  be  easily  demonstrated  by  restoring 
the  fractional  form, 

3  —    18 

"6"  —  se* 
.It  is  evident  that,  since  the  fractions  are  equal,  if  we 
reduce  them  to  the  same  denominator,   the  numerators 
must  be  equal.     The  reduction  to  the  same  denominator 
gives 

3X36  __  18X6 
6X36  "■  36X6^ 

and  because  the  numerators  must  be  equal, 

3  X  36  =  18  X  6. 

The  first  product  is  that  of  the  extremes;  the  second, 
that  of  the  means. 

8.  The  same  may  also  be  understood  by  considering 
that,  since  the  fractions  are  equal,  their  quotient  is  one; 

3. 

that  is,  77  =  1 ;   or,   by   inverting  the   second   fraction 

(XLIV.,  5),  I-  X  4?  =  !•  The  inversion  brings  the  two 
extremes  together,  and  also  the  two  means  together,  and 
proves  the  equality  of  their  products, 

9.  Proportions  have  other  properties,  which  may  be 
of  some  use  in  algebra  and  geometry,  but  of  little  im- 
portance in  practical  arithmetic. 

10.  If  any  one  term  of  a  proportion  is  unknown,  it 
can  readily  be  found  by  means  of  the  three  others ;  for, 
this  term  must  be  absent  from  one  of  the  products,  and  it 
will  suffice  to  divide  the  known  product  by  the  remaining 
factor. 

If,  for  example,  we  knew  the  three  terms,  3,  36,  and  6, 
we  could  find  18  by  dividing  the  product  3  X  36  by  6  j 
.,    ,  .     3X36        .Q 


RULE    OF    THREE.  177': 

If  it  was  6  that  was  unknown,  the  product,  3  X  36, 
divided  by  18,  would  give  6.  The  same  thing,  of  course, 
would  obtain  in  regard  to  the  other  terms,  3  and  36. 

11.  This  operation  also  has  a  distinct  name;  it  is 

THE  RULE  OF  THREE. 

The  object  of  which,  it  will  appear,  is,  hamng  given 
three  terms  of  a  proportioji,  to  find  the  fourth  term, 

12.  According  to  the  operation  just  explained. 

If  the  unknoiDn  term  is  an  extreme,  divide  the  product 
of  the  means  hy  the  other  extreme. 

If  it  is  a  mean,  divide  the  product  of  the  extremes  hy 
the  other  mean. 

13.  It  is  clear  that  there  was  no  need  of  new  expres- 
sions and  signs  for  this.  For,  having  given  as  a  condition 
the  equality  of  two  fractions, 

I-  =  i|,  for  example  : 

if  either  the  numerator  or  denominator  of  one  of  them 
was  unknown,  it  might  easily  be  found  by  means  of  the 
other  fraction. 

Let  us  suppose,  for  instance,  that  tho  numerator  18  is  not 
known.  It  is  clear,  that  since  the  second  fraction  results  from 
the  first  by  multiplying  both  its  terms  by  the  same  number,  all 
we  have  to  do  is  to  find  what  that  number  is.  This  is  done  by 
dividing  36  by  6,  and  then  multiplying  3  by  it.     Which  gives 

3  X  36  , 

,  as  above. 

6     ' 

In  like  manner,  if  the  denominator  6  was  unknown,  it  would 
be  found  by  dividing  the  known  denominator,  36,  by  the  same 
number  which  should  divide  18  to  change  it  into  3 ;  that  is,  by 
their  quotient;  and  the  operation  would  be 

,o       3X36 
36  -i-  V  '='  —T7. —  =  6,  as  above. 

•*  IS 

So  that  what  we  knew  before  was  sufficient,  without  a 
new  theory  and  new  terms. 

14.  Since  a  ratio  cannot  be  established  between  things 
which  are  not  of  the  same  nature,  it  follows  that,  in  prac- 


1*78  LESSON   XLVI. 

tical  questions,  the  first  two  terms  of  a  proportion  are  of 
the  same  nature,  dollars^  yards,  'pounds^  or  anything  else. 
The  last  two  must  likewise  be  of  the  same  nature : 

As,  when  we  say,  2  yards  cost  4  dollars ;  consequently,  6 
yards  must  cost  12  dollars :  from  which  we  form  the  proportion 
2  yards  :  6  yards  :  :  4  dollars  :  12  dollars. 

15.  Consequently,  in  the  rule  of  three,  two  of  the  three 
numbers  given  will  necessarily  be  of  the  same  nature ) 
and  the  third,  of  the  nature  of  the  number  sought. 

It  is  natural,  therefore,  to  form  the  first  ratio  with  the 
two  similar  numbers,  and  the  second  with  the  two  others. 

16.  The  Jast  being  unknown,  it  is  customary  to  repre- 
sent it  by  one  of  the  last  letters  of  the  alphabet,  x,  ?/, 
or  »•     So  that,  if  the  question  were, 

2  yards  cost  4  dollars :  what  will  6  yards  cost  ? 
The  statement,  by  proportion,  would  be  2  :  6  :  :  4  :  rr,  and 
6X4 
the  result  x  =  ~o^  =  12. 

17.  In  this  case,  we  find  that  a  greater  number  of 
yards  requires  a  greater  number  of  dollars.  The  rule  of 
three  is  then  said  to  be  direct. 

But  if  the  question  were, 

A  piece  of  work  was  done  in  8  days,  by  6  meji :  in  how 
many  days  icould  1 2  men  do  the  same  ? 

It  is  evident  that  more  men  would  require  less  days,  in 
the  ratio  of  12  to  6.    The  statement  by  proportion  would  be 

12  :  6  :  :  8  :  a:,  and  the  result  x  =  -~~  =  4. 

In  this  case,  the  rule  of  three  is  inverse.     Therefore, 
If  more  requires  more,  or  less  requires  less,  the  pro- 
portion is  direct. 

If  more  requires  less,  or  less  requires  more,  it  is  in- 
verse, 

18.  According  to  this  distinction,  in  order  to  state  the 
question  numerically  by  proportions, 

I.  Place  in  the  thii'd  term  the  number  which  is  of  the 
same  nature  with  the  answer. 


RULE    OF    THREE.  17^ 

n.  Consider,  from  the  nature  of  the  question^  whether 
the  answer  must  he  greater  or  less  than  the  third  term, 
and  then  arrange  the  first  two  numhers  in  the  same 
order  of  magnitudes. 

The  proportion  being  thus  formed,  the  unknown  term 
will  be  found  as  directed  above.  After  stating  the  opera- 
tions to  be  made  as  above, 

X  =  —7^3  for  example, 

multiply,  divide,  or  cancel,  as  is  found  most  convenient. 
The  distinction  between  direct  and  inverse  rules,  is 
of  some  use  when  proportions  are  resorted  to,  but  unne- 
cessary when  the  question  is  solved  by  simple  fractions, 
which  some  call  by  analysis.  We  will  compare  these 
two  methods  in  various  practical  questions.  For  the  pre- 
sent we  will  give  a  few  simple  exercises,  to  familiarize 
the  pupil  with  proportions. 

Questions, — Are  proportions  indispensable  in  arithmetic? 
What  is  a  ratio?  What  is  a  proportion?  How  could  it  be 
otherwise  expressed?  Give  examples.  By  what  single  name 
are  the  four  numbers  of  a  proportion  called  ?  What  is  the  first  ? 
The  second  ?  The  third  ?  The  fourth  ?  What  do  they  corre- 
spond to  in  fractions  ?  What  are  the  means  ?  The  extremes  ? 
What  relation  exists  between  them  ?  Prove  it.  If  one  term  of 
a  proportion  is  wanting,  how  can  it  be  found?  What  is  the  rule 
of  three?  Direct?  Inverse?  Give  examples.  How  do  you 
state  a  question  by  proportions  ?  How  could  any  term  be  found 
by  means  of  the  fractional  form  ?  What  is  the  ratio  of  1  to  4  ? 
Of  4  to  1  ?  Of  3  to  12  ?  Of  6  to  42  ?  Of  10  to  100  ?  Of  10  to 
1  ?  Of  15  to  5  ?  Of  6  to  3  ?  Of  12  to  48  ?  Of  48  to  12  ?  Of 
7  to  9  ?  Of  9  to  7  ?  In  a  proportion,  what  is  the  ratio  of  the 
last  two  terms  equal  to  ?  Dees  the  ratio  of  two  numbers  refer 
to  their  absolute  magnitude  ?  What  is  the  third  term  in  a  rule 
of  three  ?     How  is  the  fourth  represented  ? 

EXERCISES. 

1.  25  pounds  have  cost  650  dollars :  what  will  384  pounds  cost  ? 

A71S,  9,984  dollars. 

2.  It  took  20  days  for  135  men  to  do  a  piece  of  work :  how 
long  will  it  take  300  men  ?  Ans,  9  days, 

3.  45  men  have  built  280  yards  of  masonry  :  how  much  would 
76  men  have  built  in  the  same  time  ?  Find  the  result  in  deci- 
mals to  within  one  hundredth.  Ans,  472.89  yards. 


180  LESSON   XLVII. 

4.  If  100  men  can  finish  a  work  in  12  days,  how  many  will  do 
it  in  3  days  ?  A?is.  400. 

5.  If  3  paces  of  a  person  be  equal  to  2  yards,  how  many  yards 
will  160  paces  be  equal  to?  A?is.  lOCf. 

6.  A  garrison,  consisting  of  10,000  men,  has  provisions  at  the 
rate  of  39  ounces  a  day,  for  each  man,  for  50  days ;  3000  meu 
are  added  :  what  should  now  be  the  ration  of  each  to  last  the  same 
time?  A7is»  30  ounces, 

7.  §  of  a  yard  of  cloth  cost  2^  dollars;  what  will  Vj  yards 
cost  ?  Ans.  50  dollars, 

8.  If  7  pounds  cost  ij  dollars,  what  will  3  pounds  cost  ? 

A71S.  y^  of  a  dollar, 

9.  If  72|  yards  cost  38j  dollars,  what  will  99|  cost? 

Ans,  53  dollars, 

10.  If  0.3  of  a  house  cost  100.75  dollars,  what  will  .95  cost? 

Ans.  Sld.Oi-^- dollars. 

The  9th  question  is  a  remarkable  example  of  cancelling. 


CHAPTER  VII. 

CONTAINING   DENOMINATE   NUMBERS. 

LESSON  XLVII. 
DENOMINATE  NUMBERS. 

1.  All  the  preceding  principles  and  rules  have  had 
reference  almost  exclusively  to  abstract  numbers,  and  are 
applicable  to  all  kinds  of  units.  If  they  have  been  vi^ell 
understood,  their  application  to  denominate  numbers  will 
require  but  little  explanation. 

2.  It  has  been  said,  at  the  beginning  of  this  work,  that 
a  denominate  number  is  one  whose  unit  of  comparison  is 
designated  by  a  particular  name. 

But,  arithmetic  considers  only  denominations  which 
express  relative  numerical  values  ;  such  as  one  foot  rela- 
tively to  one  yard;  one  ounce  in  regard  to  one  pounds  &c. 


FEDERAL    MONEY.  181 

3..  So  that  smaller  units  are  fractions  of  the  larger 
ones,  but  the  numerical  relation  between  them  is  ex- 
pressed by  a  name  instead  of  a  number.  One  foot  is  one 
third  of  a  yard ;  one  ounce,  -^^  of  a  pound  ]  and,  if  the 
relative  value  of  each  inferior  unit  was  expressed  by  a 
fraction  of  the  larger  unit,  instead  of  a  name;  or,  if  the 
higher  units  were  given  as  a  whole  number  of  smaller 
ones,  the  preceding  operations  could  be  applied  at  once, 
without  any  further  explanation. 

But,  as  they  are  of  very  frequent  use,  and  their  rela- 
tions are  generally  given  in  convenient  numbers,  practice 
has  introduced  some  simple  rules,  in  combining  them, 
which  it  will  be  advantageous  to  examine. 

The  first  thing  to  be  done  is  to  commit  carefully  to 
memory  the  different  measures  and  their  subdivisions  in 
use  in  the  United  States.  For  this  purpose,  the  following 
tables  are  introduced  here. 


UNITED  STATES  CURRENCY,  OR  FEDERAL  MONEY. 

TABLE. 

Mills,    .     .     .     .     marked  m»  E.      $         d.        cts.         mills, 

10  mills  make  1  cent,     «      ct.  1  =  10  =  100  =  1000  =  10,000. 

10  cents      "     1  dime,    «      J.  1=    10=    100=     1000. 

10  dimes     "     1  dollar,  "      $.  1=      10=       100. 

10  dollars   "     1  eagle,    ''      E.  1=         10. 

This  currency  is  very  convenient ;  it  will  be  recog- 
nised as  belonging  to  the  decimal  system.  Hence,  to 
make  any  calculation  in  dollars,  it  is  only  necessary  to 
recollect  that 


Eagles 

correspond 

to 

tens. 

Dollars 

u 

units. 

Dimes 

(C 

tenths. 

Cents 

(( 

hundredths. 

Mills 

cc 

thousandths. 

and  consequently  perform  every  operation  in  Federal  money 
as  in  decimal  numbers  and  fractions. 
16 


182  LESSON    XLVir. 

The  coins  of  the  United  States  are  of  gold^  silver  and 
copper,  as  follow : 

1st.  The  gold  coins  : 

The  Double  Eagle,  worth  Twenty  Dollars. 

The  Eagle,  **  Ten 

Half  Eagle,  "  Five 

Quarter  Eagle,  *'  Two  and  a  half  Dollars. 

Three  Dollar  piece,   "  Three  ** 

Gold  Dollar,  "  One  Dollar. 

2d.  The  silver  coins — The  Dollar  =  100  cents. 
Half  Dollar  =  50  cents. 

Quarter  of  a  Dollar  =  25  cento  (frequently  called  a  Quarter). 
Dime  =  10  cents  (called  also  Ten-Cent  Piece). 
Half  Dime  =  6  cents  (called  Five-Cent  Piece ;    and  at  the 

South,  Picayune). 
The  Trirae,  or  Three-Cent  Piece. 

3d.  The  copper  coins — The  Cent;  the  Half  Cent  (now  very  rare. ) 
The  Mill  is  an  imaginary  coin,  used  merely  in  calculations  for 
greater  accuracy. 

Additional  details  relative  to  coins  and  currencies  will  be  found 
in  the  Appendix. 

NUMERATION  AND  NOTATION  OF  FEDERAL  MONEY. 

1.  From  the  decimal  relation  of  the  successive  denomi- 
nations in  Federal  money,  it  will  appear  evident,  that  any 
amount  of  that  money  might  be  written  by  placing  the 
different  denominations,  side  by  side,  in  their  order,  as  in 
common  decimal  numbers;  using  Os  in  the  place  of  absent 
denominations. 

Thus  18  eagles,  5  dollars,  6  dimes,  7  cents,  and  2  mills 
might  be  written,         E.     $.     d.     cL     m.    . 

18     5      6      7       2 
and  agreeably  to  the  remark  of  Lesson  VI. ,  7,  read  in 
various  ways )  as  for  example  ; 

185,672  mills;  18,567  cents  and'2  mills;  1,856  t^zmes  and 
72  mills;  or  again,  185  dollars  67  cents  and  2  mills,  &c., 
according  as  the  unit  of  comparison  adopted  might  be  the 
mill^  centj  dime  or  dollar. 

Likewise,  21  eagles  6  cents  and  no  mills  might  be  written 
either  21,006  cents,  if  we  wanted  the  amount  in  cents;  or 
210,060  milhj  if  the  unit  of  comparison  was  the  mill^ 


FEDERAL    MONEY.  183 

the  vacant  places  of  units  of  dollars,  cents  and  mills  being 
occupied  by  Os  as  usual. 

2.  But  this  is  not  the  customary  way  of  writing  anj 
reading  Federal  money.  The  dollar  being  its  unit  of  com* 
parison,  the  other  denominations  are  referred  to  it;  and 
accounts  kept  in  dollars,  cents  and  mills,  or  /inactions  of 
cents  instead  of  mills. 

Eagles,  though  real  coins,  do  not  appear  in  reading 
amounts  of  money,  otherwise  than  as  tens  of  dollars;  and 
likewise  dimes  are  introduced  only  as  tens  of  cents. 

Accordingly,  the  first  of  the  above  amounts  would  be 
written  $185,672     or     185.672  dollars, 

and  read  185  dollars  67  cents  and  2  millsy 

or  185  dollars  672  mills. 

The  second  would  be  written 

$210.06     that  is,  210  dollars  and  6  cents. 

The  sign  $  preceding  the  whole,  or  the  word  dollars 
following  it,  and  the  units'  point  being  placed  between  the 
dollars  and  dimes,  to  fix  the  place  of  the  unit  of  compari- 
son, THE  DOLLAR.     Hence, 

RULE. 

To  write  Federal  money,  lorite  down  the  amount  of 
dollars;  place  the  units*  (decimal)  'point  after  it;  and 
next  to  this  ivrite  the  cents  and  mills  (or  fractions  of  cents 
instead  of  mills);  Fill  the  places  of  absent  denominations 
hy  zeros;  and  put  the  sign  $  before  the  number y  or  write 
the  word  dollars  after  it. 

Thus  2  dollars  and  5  cents  would  be  written     $2.05 
3       ''      and  7  mills  "  $3,007 

6i  cents  referred  to  the  dollar  $0.06}  or  $  .06} 
5  cents  and  2  mills     do.  $0.05    or  $  .052 

9  mills  do.  $0,009  or  $  .009 

RULE. 

3.  To  read  Federal  money,  read  the  number  to  the  left 
of  the  units*  point  as  dollars;  the  next  two  figures  as  cents, 
and  the  third  as  mills 

Dimes  are  not  named  in  the  reading  of  Federal  money, 


184  LESSON    XLYII. 

and   it  is  customary  to  change  them  into  cents  bj  the 

addition  of  a  0 ;  thus : 

;~  .  2  dollars  and  4  dimes — that  is,  decimally,  82.4 

is  written  in  preference  $2.40 

which  is  the  same  thing;  and  reads  2  dollars  and  40  cents, 
or  also  240  cents. 

4.  Since  the  figures,  which  compose  an  amount  of  Federal 
money,  are  decimally  connected,  it  follows  (VL,  7,)  that  it 
may  be  read,  not  only  in  the  usual  way  as  dollars,  cents 
and  7nills,  but  also  as  cents  and  mills,  or  as  all  mids'j  thus: 

$2,654  may  be  called,  2  dollars,  65  cents  and  4  mills, 
265  cents  and  4  mills , 
or  even  2,654  mills. 
For  the  same  reason  2  dollars  or  $2.00 

may  be  called  200  cents 

and  also  2000  mills 

using  in  place  of  the  vacant  denominations  additional  zeros, 
which  we  know  do  not  alter  the  value  of  numbers  after 
decimal  fractions,  (XXL,  8);  for  they  do  not  change  the 
relative  position  of  the  denominations. 

5.  Very  frequently,  especially  in  mercantile  business, 
fractions  of  cents  are  used  instead  of  mills,  as  for  instance, 

$10. 12  J    which  reads  ten  dollars  12  J  cents, 

0.18}  "  181  cents, 

25.33|  ^'  twenty-five  dollars  33J  cents. 

Sometimes  the  cents  are  written  as  vidgar  fractions  of 
dollars  in  smaller  figures,  as  $10j^;  ^^^iS^  ^^^^5^^  ^^^ 
often  also  in  the  bills  of  merchants  simply 
$101i^;     $25?H;     $26511. 
The  last  mode,  however,  is  liable  to  cause  errors. 

In  deeds,  drafts  and  important  documents  in  general, 
amounts  of  money  are  written  at  full  length  in  words  :  In 
that  case,  it  is  a  good  practice  to  introduce  also  the  num- 
bers in  figures  between  brackets,  as  an  additional  security 
against  errors. 

In  accounts  and  bills,  the  decimal  point  is  not  used : 
The  dollars  and  cents  are  separated  in  two  different  columns 
by  a  vertical  line,  as  in  the  following  e:iample. 


FEDERAL    MONEY. 


185 


Wm.  Smith,  to  Ths.  Jones, 


Richmond,  August  25th,  1856. 
Dr. 


Jan'y  6th 
March  7 

April  20 
July  9 


To  25  pounds  of  Sugar  at  12J  c 
*'  2         "  Tea      "  1.15 

"  2         "  Coffee  *'     .18| 

*'  20       *'  Bacon  "     16f 

*'  1-|-  gallon  of  Wine    «*  $325 

Rec'd.  Paym't.  Sept.  1st  1856.  $  13      197_ 

Ths.  Jones  by  A.  Grant,  cl'k.  12 


(^Cheque  to  pay  for  the  above.) 

Bichmond,  Sept.  1st,  1856. 

Cashier  of  the  Bank  of  Virginia  pay  to  Ths.  Jones  or 
bearer  Thirteen  -^^jj  dollars. 

^13.fg%*  Wm.  Smith. 

Questions. — What  are  denominate  numbers?  What  kind  of 
denomination  does  arithmetic  consider  ?  Give  examples.  Could 
these  names  be  expressed  by  numbers  ?  By  what^nd  of  num- 
bers would  you  express  inferior  units?  By  what  kiH  of  numbers 
would  higher  units  be  expressed  in  regard  to  lower  ones  ?  What 
is  Federal  money  ?  Repeat  the  table.  What  are  the  gold  coins  ? 
The  silver  coins?  The  copper  coins  of  the  United  States?  How 
are  amounts  of  Federal  money  written  ?  How  read?  How  are 
cents  written  in  fractional  form  ? 


EXERCISES. 

1.  Read,  as  dollars,  cents,  and  mills,  $18.30;  $55,367;  $5.4. 

2.  Read,  as  cents,  $16.51;  $3.40;  .07. 

3.  Write  fourteen  dollars  sixteen  cents  and  three  mills. 

4.  **      seven  dollars  seven  cents  and  seven  mills. 

5.  "      six  dollars  and  six  mills. 

6.  "      twenty  dollars  and  four  cents. 

7.  **      six  cents  and  a  quarter. 

8.  "      twelve  and  one  half  tents. 

9.  *'      eight  and  two  thirds  cents. 

10.  *'      thirty-one  and  one  fourth  cents. 

11.  "      ten  dollars  twenty-five  cents. 

12.  *'      one  hundred  and  six  cents. 

13.  "      two  hundred  and  thirty-one  cents  and  a  quarter. 

-  It  is  customary  to  omit  fractions  smaller  than  J;  but  to  add  one 


where  they  exceed  it; 
1  in  accounts. 

16* 


as  10  J,  it  is  alternately  omitted  and  made 


1S6  LESSON    XLVII. 

14.  In  35  dollars  and  9  mills,  how  many  mills? 

15.  In  6  dollars  25  cents  and  1  mill,  how  many  mills? 

IG.  In  $150,  how  many  cents  ?     How  many  mills?     How  many 
dimes  ? 

17.  How  many  mills  in  $9.67;  in  $6.5;  in  $5;  in  75  cents; 
in  9  cents? 

18.  How  many  cents  in  $500.29  ?     In  60.2  ? 

II.   ADDITION,  SUBTRACTION,  MULTIPLICATION,  AND  DI- 
VISION OF  FEDERAL  MONEY. 

6.  The  dollar  being  the  luiit  of  comjmn'son  of  Federal 
money,  dimes  are  tenths  of  that  unit, 

cents  "  hundredths^ 
mills  '^  thousandths^ 
and  consequently,  addition,  gUBTRACTiON,  multiplica- 
tion and  DIVISION  in  Federal  money  are  mere  operations 
in  decimal  fractions,  which  were  thoroughly  explained  in 
Chapter  111.,  and  therefore  a  few  examples  will  suffice  in 
this  place.  . 


^        EXAMPLES   IN   ADDITION. 

7.  I.  If  it  was  proposed  to  add  $13 
and  25  cents;  $105  and  6  cents;  $78 
and  15  mills;  33  cents;  the  operation 
would  be  arranged  and  performed  as 
an  ordinary  addition,  as  may  be  seen 
in  the  annexed  operation. 


OPERATION. 

$13.25 
105.06 
78.015 
.33 

$196,655 


II.  The  bill  in  No.  5  is  an  example  of  addition  with 
fractions  instead  of  mills:  The  fractions  are  added  first, 
and  give  ^^  and  1  to  be  carried.  The  rest  of  the  operation 
as  usual. 


EXAMPLES   IN    SUBTRACTION. 


8.  I.  Let  it  be  proposed,  for  instance, 
to  subtract  56  dollars  25  cents  and  9 
mills  from  105  dollars  19  cents. 

The  operation  would  be  as  shown 
here. 


OPERATION. 
105.19    0 

56.25  9 
$48.93  1 


FEDERAL  MONEY. 


187 


II.   The  annexed  operation  is  an  example  with  fractions 

of  cents.  OPERATION. 

From  82,014.124 

To  subtract        927.661 


3 

—  4 
6 


§l,086.45g 

Since  |  or  |  is  greater  than  1  or  |,  we  borrow  1=|,  from 
which  I  is  subtracted,  leaving  |,  which  added  to  the  upper 
I  (Lesson  XLIL,  9),  makes  |;  this  we  set  down  and  carry 
1  to  the  lower  figure  6;  1  and  6  are  7,  and  7  from  12  leaves 
5  and  1  to  carry,  &c.     The  rest  of  the  operation  as  usual. 

III.    From  $110  to  subtract  6?  cents. 

In  such  a  case  it  is  better  to  equalize        operation. 
the    denominations   as    shown  here,  by           $110.00 
the  addition  of  Os,  though  it  is  not  in-                 0.06 J 
dispensable,  (XXII.)  '  

^  ^  $109,931  . 


EXAMPLES   IN  MULTIPLICATION. 

9.  I.  The  multiplier  a  lulwle  niimhcr.  Let  it  be  pro- 
posed to  multiply  25  dollars  56  cents  and  8  mills  by  12. 

It  is  clear  that  the  multiplication  of  7niUs  and  cents  will 
give  respectively  mills  and  cents,  just  as       operation. 
the  repetition  of  thousandths  and  him-  $25,568 

clredths  gives  thousandths  and  hundredths  12 

in  the  product;  therefore  we  multiply  as 

in  decimal  fractions,  and  cut  off  by  the         $306,816 
units'  point  as  many  denominations  or 
decimals  as  are  contained  in  the  multiplicand. 

II.  The  multiplier  containing  decimal  and  other  frac- 
tions.    For  instance : 

A  bookseller  orders  books  for  an  individual,  and  to  cover 
all  expenses  is  to  get  $1.37i  to  the  dollar  on  the  invoice, 
which  amounts  to  $216.622 — how  much  is  he  to  receive? 

This  multiplication  may  he  performed  in  two  different 
ways — either  by  retaining  the  fractions  of  cents,  or  more 
simply,  hy  cha7iging  them  to  decimals,  that  is,  in  this  case, 
mills. 


1«8 

LESSON 

XLYII. 

OPERATION. 

(1) 

(2)  Changed  to  decimals 

$216.62^ 

$216,025 

1.37| 

1.875 

product  of  J 

by  137  J 

68J 

1083125 

product  by  J 

10831 

1510375 

"       by  7 

151634 

649875 

by  3 

64986 

216025 

"       by  1 

21662 

$297.85931 


$297.859375 


In  either  case,  cut  off,  according  to  the  rules  in  decimal 
fractions,  (Lesson  XXV.,  3),  as  many  decimal  figures  as 
there  are  in  both  the  multiplicand  and  multiplier.  The 
result  is  297  dollars  85  cents  9  mills  and  -f^^j^-=^  of  a  mill. 

Recollect  that  the  multiplier  1.37  J  is  considered  as  an 
atetract  number. 


EXAMPLES   IN   DIVISION. 

10.  Since,  in  our  mode  of  proceeding  in  the  division  of 
decimals,  we  make  the  divisor  a  whole  number  by  striking 
off  the  units'  point  (XX VT.,  2),  we  will  at  once  consider 
the  case  of  division  by  decimal  fractions. 

Question.  140  dollars  90  cents  61  mills  have  been  paid 
for  41  f  yards  of  cloth  ;  how  much  is  it  a  yard  ? 

The  fractions  being  written  decimally,  the  operation  will 


OPERATION. 


§140/90.625 
15  658 
3  1312 

20875 


n/75. 

3.375 
=:S3.37i 


be  as  follows  by  quick  division  : 

we  strike  off  the  decimal   point 

in  the  divisor  to  make  it  a  whole 

number,  and  remove  that  of  the 

dividend   the   same   number  of 

decimals  to  the  right;  and  thus 

see  at  once  that  the  result  must  be  a  number  of  thousandths 

or  mills ;  so  that  the  result  is  3  dollars  37  cents  5  mills. 

11.  If  the  fractions  were  not  conveniently  transformed  into 
decimals,  and  perfect  accuracy  was  required,  the  division 
should  be  made  as  in  mixed  numbers  (XLIV.,  8).  But 
this  is  rarely  the  case,  and  the  change  of  small  fractions  to 
decimals  furnishes  generally  a  sufficient  approximation. 


FEDERAL    MONEY.  189 

EXERCISES    IN    THE    FOUR    RULES. 

1.  Multiply  375  dollars  and  6  cents  by  62.     Ans.  $23,253.72. 

2.  From  296  dollars  4  cents  2  mills  take  117  dollars  59  cents 
6  mills.  Ans.  $ 

3.  From  ten  dollars  and  6  mills  take  9  mills.  Ans.  $ 

4.  Bought  338  sheep  at  2  dollars  69  cents.  ^7^5.  $909.22. 

5.  What  will  89J  yards  cost  at  34  cents  6  mills  per  yard  ? 

Ans.  $30.88_i^. 

6.  Bought  96  yards  of  cloth  at  5  dollars  67  cents  per  yard. 

Ans.  $544.32. 

7.  From  one  hundred  dollars  take  65  cents.  A?is.  $ 

8.  Bought  484  lbs.  of  tallow  at  7  cents  3  mills.  Ans.  $35.33.2. 

9.  From  one  thousand  dollars  take  sixteen  cents.    Ans.  $ 

10.  How  much  are  315.25  cubic  yards  of  building    stones  at 
$8.62J  a  yard?  A7is.  $2719.031J. 

11.  Divide  276  dollars  75  cents  by  12.  Ans.  $23,062^. 

12.  Divide  2754  dollars  by  216.  Ans.  $12.75. 

13.  Sold  48  bales  of  cotton,  each  containing  397  pounds,  at  13 
cents  7  mills  per  pound.  Ans.  $2,610.67.2. 

14.  A  man  received  51  dollars  and  52  cents  for  16  days.     How 
much  did  he  earn  a  day?  Ans.  $3.22. 

15.  Bought  7J  bales  of  cotton,  each  327  pounds,  at  $9.37^  cts. 
per  hundred  pounds;  what  is  the  whole  cost?  $244.851_9_, 

N.  B.    In  such  questions  observe  that  the  division  by  100  pounds 
gives  decimals  in  the  multiplier. 

16.  A  railroad  has  cost  $159,799.89  for  12|  miles;  how  much 
is  it  per  mile?  A?is.  $12,533.32  4||. 

17.  What  will  be  the  cost  of  carrying  459.75  cubic  yards  of 
earth  1569  feet  at  9  mills  per  yard  for  100  feet?  Ans.  $64,92. 1-f 

18.  There  was  paid  $1,433.16  to  27  men,  how  much  was  it  to 
each  man?  Ans.  $53.08. 

19.  What  is  the  cost  of  832  bushels  of  wheat  at  1  dollar  18J 
cents  per  bushel  ?  Ans.  $988. 

20.  How  much  are  357  pounds  of  coffee  at  13|  cents  ?  Ans.  $ 

21.  How  much  are  18|  yds.  of  calico  at  25  cts.  ?    Ajis.  $4.68|. 

22.  How  much  are  125|  acres  of  land  at  $18.75? 

Ans.  $2,357.81}. 

23.  Bought  a  farm  of  266J  acres  for  $8,115.65,  how  much  is  if 
an  acre?  Ans   $30.47.2  (nearly). 

24.  What  will  96.675  feet  of  board  cost  at  $11.65  a  thousand  ? 

Ans.  $ 

25.  What  will  376  75iL  M.  of  bricks  come  to  at  $6.87^  per  M.  ? 
(M.  means  1000  bricks.)  Ans. 

26.  What  will  17,625  ft.  of  plank  cost  at  $12.75  per  thousand? 

Ans.  $224.718f. 

27.  Bought  42  barrels  of  apples,  each  containing  3  bushels,  for 
$39. 37 J  ;  how  much  is  it  a  bushel?  Ans.  $ 


190 


LESSON    XLVIII. 


28.  Bought  12  lbs.  coffee  at         10^  cts.  a  pound 
4    "     tea        '«  $1.31^ 
00    '<     sugar    "         12.^         " 
15    *'     cheese"         ll''         " 
105-^    '*     butter  "         17  " 


4  gallons  wine  <*  $4.25  a  gallon 

29.  Sold  28  yds.  linen         at       62^-  cts. 

8    ''     calico         "        18|    '* 

24    '*     gingham    "        37.^     " 

11     "     silk            ♦♦  $1.18|    ** 

J    "    vesting      *'  $3.00      " 

$51,335 

$ 

$43.31J. 
LESSON  XLVIII. 
ENGLISH  CURRENCY,  OR  STERLING  MONEY. 


TABLE. 

£,     sh,      d,       far, 

1:=  20=  240  =  960. 

1=    12=    48. 

1=      4. 

Farlhings  are  ge- 
nerally expressed  by 
fnictions ;  ^d.  for  1; 
i  for  2  J  i  for  3. 

Note. — This  table  is  important,  on  account  of  the  great  inter- 
course with  England,  and  because  it  was  used  in  this  country 
before  the  revolution.  It  is  still  used  by  some  people,  but  the 
value  of  the  pound  varies  in  different  states. 

[Where  one  I  And  one 


2  farthings  make  1  halfpenny,  marked  \, 

4  farthings      "     1  penny,  "       d, 

12  pence    '        "     1  shilling,  "     sh, 

20  shillings       "     1  pound  sterling,"      £, 

21  shillings       "     1  guinea. 


Thus,  one  dollar  is  5  sldllings  in  Canada  and 

Nova  Scot 


8.5A. 

10*/i. 

65/i. 


nd  \ 
ia,  \ 

in  New  York  and  Ohio, 

in  North  Carolina, 

in  New  England  States,  ^ 

Virginia,  ! 

Kentucky,  | 

Tennessee,  J 

"       Ish,  6^.   in  New  Jersey,  ^ 

Pennsylvania,  I 

Delaware,  j 

Maryland,  J 

"       ^sh,  Sd.  in  South  Carolina  and  ) 

Georgia.  J 


hilling  is 
20    cents 

22  cents 
10    cents 

16|  cents 


pound 
$4.00 

$2.50 
$2.00 

$3.33:|- 


13}  cents 


21-1- cents 


$4f 


WEIGHTS,    MEASURES,    ETC.  19X 

Federal  money  is  rapidly  driving  away  this  troublesonie  va- 
riety of  currencies. 

TROY  WEIGHT. 

Grains,       .     .     .  marked     .     .     gr.  lb,   oz.     dwt.          gr* 

24  grains  make       1  pennyweight,   divt.  1  =  12  =  240  =-=  5,760. 

20  pennyweights    1  ounce,                  oz,  1=    20=     480. 

12  ounces                1  pound,                  lb.  1  =       24. 

B}^  this  weight  precious  metals  and  jewels  are  weighed. 
Troy  weight  is  also  frequently  used  by  chemists. 

AVOIRDUPOIS  WEIGHT. 

Drams,         marked      dr. 

16  drams  make     1  ounce,  "  oz, 

16  ounces  "       1  pound,  "  lb, 

28  pounds  "       1  quarter,  "  qr, 

4  quarters  "       1  hundred  weight,     "  civt, 

20  hundred  weight "       1  ton,  "  ton  or  T, 

T.     cwt,    qr,         lb.  oz.  dr. 

1  =  20  =  80  =  2,240  =  35,840  =  573,440. 

1=    4=     112=    1,792=    28,672. 

1=       28=       448=      7,168. 

1=         16=         256. 

1=  16. 

By  this  weight  are  weighed  all  coarse  articles:  such 
as  hay,  grain,  groceries,  and  all  baser  metals. 

Note. — 1 /5.  avoirdupois  =  14  o;2r.  Wdwt.  wigr.Troy, 
loz,         «  =    0         18  5^-. 

l^r.         «  =0  1  3^. 

The  pound  avoirdupois  contains     6, 999 J  grains. 
The  pound  Troy,  U.  S.  Standard,  5,760         '* 
175  Troy  pounds  are  equal  to  144  avoirdupois. 
175  Troy  ounces  =  192  avoirdupois. 

In  some  states,  the  hundred  weight  is  reckoned  only  100  lbs., 
and  the  ton  2,000  lbs.     This  is  certainly  preferable. 

In  England,  the  standard  of  weight  is  obtained  from  a  cubic 
inch  of  water  which  weighs  252.458  grains;  of  such  grains  5.,760 
make  a  pound  Troy,  and  6, 999 J  a  pound  avoirdupois. 


192  LESSON    XLVIII. 

APOTHECARIES'  WEIGHT. 

Grains,         .         .         marked  gr.     lb 
20  grains  make  1  scruple,  "  sc.  or  J).  1 : 

3  scruples  "     1  dram,      "  dr.  or  ^. 

8  drams      "     1  ounce,     "  o^.  or,^. 
12  ounces     "     1  pound,     "  lb.  or  lb. 

Apothecaries  and  physicians  make  use  of  this  weight  in  com- 
pounding medicines;  but  they  buy  and  sell  their  drugs  by  avoir- 
dupois weight. 

The  pound,  ounce,  and  grain  are  the  same  as  in  Troy  weight. 

LONG  MEASURE. 


o 

12  = 

3 

96  = 

9 

288  = 

gr, 
5,760. 

1  = 

8  = 

24  = 

480. 

1  = 

3  = 

60. 

1  = 

20. 

3    barley  corns         make  1  inch. 

marked 

in. 

12    inches     .         .         " 

1  foot, 

« 

ft. 

3    feet          .         .         " 

1  yard, 

a 

yd. 

5]  yards  or  16^-  feet    « 

1  rod,  perch,  or 

pole,     " 

rd. 

40  "rods         .         .         " 

1  furlong. 

cc 

fur. 

8    furlongs  or  320  rods" 

1  mile, 

ii 

mile,  TTil, 

3    miles       .          .         " 

1  league, 

a 

lea.  or  L, 

60    geographical,  or  )    ^^ 
69  y^  statute  miles  $ 

1  degree, 

ti 

dcg.  or  ". 

360  degrees  the  circumference  of  the  earth. 

vy.ile.  fur.    rd. 

yds.        ft. 

in. 

1  =  8  =  320  : 

=  1,760=  5,280  = 

:  63,360. 

1=    40  = 

=     220=     660  = 

7,920. 

1  = 

=       5}=     16^  = 

198. 

1=3=         36. 
1    =         12. 

A  fathom  is  six  feet,  and  is  used  to  measure  the  depth  of  water. 

A  hand  is  four  inches,  and  is  used  to  measure  the  height  of 
horses. 

One  span  is  nine  inches. 

Six  points  are  one  line. 

Twelve  lines,  one  inch :  sometimes  the  inch  is  subdivided  into 
ten,  sometimes  into  eight  lines. 

The  standard  yard  has  been  fixed  in  England  by  the 
length  of  the  pendulum  vibrating  seconds;  in  London,  at 
a  temperature  of  62*^  Fahrenheit,  the  yard  is  ^fy'.fff  of 
the  pendulum^  which  is  39.1393  inches  in  length. 


WEIGHTS,    MEASURES,    ETC. 


193 


CLOTH  MEASURE. 


2i  inches 

make  1  nail. 

marked 

nl. 

4    nails 

1  quarter  of  a  yard 

qr. 

4    quarters 

1  yard. 

yd. 

3    quarters 

1  ell  Flemish,      . 

E,Fl. 

5    quarters 

1  ell  English, 

E,  E. 

4    quarters  li 

inches     " 

1  ell  Scotch, 

E.S. 

6    quarters    . 

yd,    qr 

1  =  4 

1 

1  ell  French, 
nl,      in, 
=  16=  36. 
=    4=    9. 
1=    2|. 

E.Fr, 

This  measure  is  used  for  all  kinds  of  stuffs ;  the  Eng- 
lish, Flemish,  &€.,  are  used  for  goods  imported  from  the 
respective  countries  from  which  they  take  their  names. 


LAND  OR  SQUARE  MEASURE. 

144 

9 
30^ 
40 

4 

inches          .     make     1  square  foot,    marked      . 
square  feet           "         1  square  yard,          " 
square  yards        «         1  square  pole,           " 
square  poles        <'         1  rood,              •         " 
roods           .         *'         1  acre,              .         '^ 

sq.ft. 
sq,  yd, 
P. 
R. 
A. 

640 

acres           .         " 

1  mile,             .         " 

M. 

A,    R,     P, 
1=4=  160  = 

sq,  yd,       sq.ft,          sq.  ^?^. 
4,840  =  43,560  =  6,272,640. 

1=    40  = 

1,210=  10,890=  1,568,160. 

1  = 

30|-=     272i=       39,204. 

1    =          9    =          1,296. 

1    =             144. 

By  this  measure  land  and  all  surfaces  are  measured ; 
jsuch  as  boards,  glass,  pavements,  plastering,  &c. 

In  measuring  land,  a  4-pole  chain,  or  Q^  feet  in  length, 
IS  used.  It  is  divided  into  100  links,  each  of  v^hich  is 
7.92  inches  in  length. 

10  square  chains  make  1  acre. 

There  are  640  acres  in  a  square  mile. 

A  square  inch  has  its  side  1  inch  long. 

A  square  foot  has  its  side  1  foot  long. 
17  N 


194 


LESSON    XLVIII. 


A  square  yard  has  its  side  1  yard  long. 

The  difference  between  3  square  ^  square  feet. 

feet  and  ^  feet  square^  is  explained 
by  the  annexed  diagram. 

3  square  feet  means  3  squares, 
each  of  which  has  its  four  sides  one 
foot  long. 

3  feet  square  means  a  square 
whose  sides  are  3  feet  long.  It 
contains,  as  the  figure  shows,  9 
square  feet ;  and,  in  general,  a 
square  contains  a  number  of  square 
feet  equal  to  the  product  of  its  side 
by  itself. 

So- that  12  feet  square  =  144  square  feet ;  and  the  num- 
ber of  small  squares  contained  in  a  square  figure,  is  equal 
to  the  product  of  its  two  sides.  Thus,  a  figure  4  feet  in 
length  and  3  feet  in  breadth,  would  contain  12  square  feet. 


3  feet  square. 

1  ton. 


1  cord  of  wood, 


C.ft. 
C.yd. 


c. 


SOLID  OR  CUBIC  MEASURE. 
1,728  solid  inches        .         make     1  cubic  foot,    marked 
27  solid  feet  ,  "        1  cubic  yard,         " 

40  C.  feet  of  round,  or      K^ 
50  C.  feet  of  hewn  timber  J 
128  feet,  solid  feet;  that  is,  j 
a  pile  8  feet  long,  4  feet  \  " 
wide,  and  4  feet  high     \ 
4  feet  wide,  4  feet  high,  and  1  foot  in  length,  make  1  cord  foot 
of  wood. 

By  this  table  the  solid  contents  of  bodies  are  deter- 
mined; such  as  stone,  timber,  earth,  boxes  of  goods,  &c. 

A  cube  is  a  solid,  having  all  its  sides  equal  and  its  faces 
squares. 

1  cubic  inch  has  sides  1  inch  long. 

1  cubic  foot         "         1  foot  long  =  1,728  inches. 

1  cubic  yard        "         1  yard  long  =  27  cubic  feet. 

In  general,  solid  contents  are  the  number  of  cubic 
units  formed  by  the  product  of  the  height  by  the  breadth 
and  thickness.  Thus  it  is  that  a  cord  is  8x4x4  =  128 
cubic  feet. 


WEIGHTS,    MEASURES,    ETC. 


195 


Masonry  is  also  measured  by  the  perch,  which  is 
1  perch  in  length,  1|-  feet  in  breadth,  and  1  foot  high ; 
that  is,  161 X  1^x1  =  24.|  cubic  feet.  The  cubic  yard 
should  be,  and  is  very  frequently,  preferred. 


WINE  MEASURE. 

This  is  used  for  all  liquors,  except  beer,  ale  and  milk. 
The  U.  S.  Standard  gallon  contains  231  cubic  inches. 


Gills, 

marked 

gi' 

4    gills 

make 

1  pint,          " 

pt. 

2    pints 

a 

1  quart,        " 

qt. 

4    quarts 

c: 

1  gallon,       « 

gal. 

31 2"  gallons 

ec 

1  barrel,       « 

hi. 

63    gallons        '    , 

a 

1  hogshead,  " 

hhd. 

2    hogsheads 

a 

1  pipe,          « 

pi. 

2    pipes      .         •             « 

1  tun,            « 

tun* 

tun.    pi.  hhd.  hi.   gal. 

qt.            j)t.             gi. 

1=  2=4=8=252 

=  1,008=2,016=  8,064. 

1=2=  4=  126 

=     504=  1,008=4,032. 

1=2=    63 

=     252=     504=2,016. 

1=    3l| 

=     126=     252=1,008. 

1 

=         4=         8=       32. 

1=         2=         8. 

1=         4. 

This  gallon  weighs  8.3389 
foot. 

pounds.     There  are  7.48  in  a  cubic 

Besides  these, 

42  gallons  make       .     make     1  tierce,     .     marked 

tier. 

84  gallons     « 

a 

1  puncheon,         " 

•pun. 

ALE  AND  BEER  MEASURE. 

This  is  used  in  measuring  ale,  beer,  and  milk. 

marked 
make       1  quart,         .         ^^ 


Pints, 
2    pints 
4    quarts 
36    gallons 
l|-  barrels  =  \ 
54  gallons  J 
2    barrels 
2    hogsheads 
2    butts 


1  gallon, 
1  barrel, 

1  hogshead, 

1  puncheon, 
1  butt, 
1  tun. 


pt, 
qt, 
gal, 
bar, 

hhd. 
picn, 
hutt, 
tun. 


196 


LESSON   XLVIII. 


htttt.  •pun,  hhd.  bar,     gal, 
1=  1^=2    =  3    =  108: 
1    =1|=2    = 

1  =  u  = 


1    = 


qt.        pt. 

432  =  864. 
72=288=  576. 
54=216=  432. 

:28S. 


36  =  144  = 
1=      4=      8. 
1=      2. 

A  gallon,  beer  measure,  contains  282  cubic  inches ;  and  it  is 
remarkable  that  the  wine  and  ale  gallons  have  the  same  propor- 
tion to  each  other  as  the  troy  and  avoirdupois  pound. 

Besides  the  above,  there  is  the  kilderkin,  vv^hich  is  -|  barrel, 
and  the  firkin,  which  is  \  barrel. 

DRY  OR  CORN  MEASURE. 
This  is  used  in  measuring  all  dry  articles. 


make 


Pints, 
2  pints 

4  quarts        •  *' 

2  gallons  or  8  quarts  " 
4  pecks         .  " 


marked 


pt. 

qt. 
gal. 
pk. 
lu. 
ch. 


1  quart,  " 

1  gallon,  " 

1  peck,  " 

1  bushel,  « 

36  bushels  «         1  chaldron,       " 

ch,     hu,      ph.      gal.        qt.  pt. 

1=  36=  144=  288=  1,152=  2,304. 
1=      4=      8=       32=       64. 
1=      2=         8=        16. 
1=  4=         8. 

1=         2. 
The  gallon,  dry  measure,  contains  268y  cubic  inches.    A  Win- 
chester bushel  is  18^  inches  in  diameter,  and  8  inches  deep ;  it 
contains  2,150^  cubic  inches.     It  is  the  United  States  Standard, 

But  the  coal  bushel  must  be  19|-  inches  in  diameter  :  36otthese 

bushels  make  a  London  chaldron  of  coal,  the  weight  of  which  is 

3,156  pounds,  or  nearly  1  ton  8  cwt. ;  and  a  bushel,  8S  pounds. 

The  barrel,  for  measuring  unshelled  Indian  corn,  contains  5 

bushels. 

Besides  the  above, 
2  quarts      make         .         .         1  pottle,      .         marked    pot, 
8  bushels         "  .         .         1  quarter,  "  qr. 

5  quarters        "  .         .         1  wey  or  load,         "  ii-eij. 

2  weys  "  .         .         1  last,  .  "  last. 

By  an  act  of  parliament,  passed  in  1824,  and  carried  into  exe- 


WEIGHTS,    MEASURES,    ETC. 


197 


cution  1st  January,  1826,  a  greater  uniformity  has  been  estab- 
lished ;  and  now  only  one  gallon  measure,  whether  for  grains  or 
liquors,  is  allowed  in  the  United  Kingdom.  This  standard  im 
perial gallon  contains  277.274  cubic  inches  ;  its  weight,  10  pounds 
of  water,  at  62^. 

8  such  gallons  make  the  bushel,  which  is  2218.192  cubic  inches. 

The  United  States  gallon  contains  9.7034  pounds  of  water, 
avoirdupois :  a  cubic  yard  contains  21.69  bushels. 

TIME. 


Seconds, 

• 

markt 

jd     .         sec,  or 

60  seconds 

make 

1  minute,             " 

m.  or ' 

60  minutes 

a 

1  hour,                  " 

hr. 

24  hours 

({ 

1  day,                   « 

day. 

7  days 

(( 

1  week,                " 

wh. 

30  days 

(( 

1  month,              " 

mo. 

52  weeks  1  day  6  hours,  (^ 

1  common   or  (  jj 
Julian  year,  \ 

yr. 

or  365± 

days,  make     ) 

yr.    mo,    wh,     day. 

hr,            m. 

sec. 

1=12=52=365^  = 

8,766=525,960  = 

31,557,600. 

1=      7    = 

168=    10,080  = 

604,800. 

1    = 

:       24  =      1,440  = 

86,400. 

1=           60  = 

3,600. 

* 

1  = 

60. 

The  exact  solar  year  is  365  days  5  hrs.  48'  48'^  In 
reckoning  time,  only  the  365  days  are  counted  as  one  year: 
the  odd  six  hours,  by  accumulating  for  4  years,  make  one 
day,  which  is  then  added,  and  gives  366  days  to  every 
fourth  year,  called  bissextile  or  leap  year. 

This  additional  day  is  given  to  the  month  of  February, 
and  makes  it  29  days  long,  in  every  year  v^^hich  is  a 
multiple  of  4. 

CIRCULAR  MEASURE. 
This  is  used  in  geometry,  astronomy,  and  geography. 
Seconds, marked    ". 


60  seconds 

make  1  minute, 

iC 

', 

60  minutes 

"     1  degree. 

t( 

o^ 

30  degrees 

12  signs  or  360^ 

"     1  sign, 

"     1  circumference  of  a  circle, 

S, 
Cir, 

198 


LESSON   XLVIII. 


Cir,   5.       °  ' 

1=  12=  360  =  21,600=  1,296,000. 

1=    30=    1,800=     108,000. 

1=         60=         3,600. 

1=  60. 


MISCELLANEOUS 

12  things               .  make 

12  dozen                .  " 

12  gross  or  144  dozen  '' 

20  things                .  « 

112  pounds               .  " 

24  sheets  of  paper  " 

20  quires                .  '' 

56  pounds  of  flour  '' 

196  pounds  of  flour  ** 

200  pounds  of  salt  meat  *' 

A  sheet  folded  in 

2  leaves,  which  make  4  pages, 

4       a            a            a       g  (( 

8       «            «            "     16  " 

12      «          «          «    24  « 

18      «          "          "    36  *« 


TABLE. 

1  dozen. 

1  gross. 

1  great  gross. 

1  score.  • 

1  quintal  (of  fish). 

1  quire. 

1  ream. 

1  bushel. 

1  barrel. 

1  barrel. 

a  folio. 

a  quarto  or  4to. 

an  octavo  or  8vo. 

a  duodecimo  or  12mo. 

an  octodecimo  or  18mo. 


REMARKS   ON    THE    PRECEDING    TABLES. 

I  have  not  divided  these  tables  into  lessons,  because 
they  are  to  be  used  chiefly  for  reference ;  nor  do  I  think 
it  necessary  to  set  down  questions.  The  tables  have  to 
be  committed  to  memory,  and  the  questions  will  suggest 
themselves  readily  to  the  teacher. 

It  will  be  observed  that  the  scale  between  the  several 
denominations  is  very  irregular,  though  the  ratios  are 
generally  simple  numbers.  It  would  be  a  great  deal 
better  if  the  decimal  scale  of  divisioa  could  have  been 
adopted  for  all. 

There  is,  however,  an  advantage  in  having  distinct 
names  for  the  successive  fractional  subdivisions  of  the 
same  unit. 


TRANSFORMATION  OF  DENOMINATE  NUMBERS.    199 

If  we  were  to  use  only  the  smaller  unit,  we  would 
frequently  have  occasion  to  employ  very  large  numbers. 

If,  on  the  contrary,  we  were  to  refer  numerically  the 
subdivisions  to  the  largest  unit,  calculations  and  the  opera- 
tions of  commercial  business  would  be  encumbered  with 
fractions. 

Even  in  Federal  Money,  where  the  regularity  of  the 
decimal  system  is  introduced,  each  subdivision  of  the  dol- 
lar, for  this  reason,  has  a  special  name  different  from  the 
abstract  scale  of  numeration. 

Questions, — Is  the  scale  between  the  different  subdivisions  of 
the  same  unit  regular?  Would  the  decimal  scale  have  any  ad- 
vantage? Is  there  any  advantage  in  giving  different  names  to 
the  successive  subdivisions  of  the  same  unit  ?  And,  if  so,  what 
are  they?  In  the  decimal  system  of  federal  money,  is  the 
change  of  name  an  advantage  ? 

LESSON  XLIX. 

TRANSFORMATION  (reduction)  OF  DENOMINATE 
NUMBERS. 

1.  In  operations  with  denominate  numbers,  it  is  fre- 
quently advantageous  to  express  a  number  of  units  of  a 
certain  denomination  either  in  a  greater  number  of  rela^ 
tive  units  of  a  lower  denomination^  or  in  a  smaller  num^ 
her  of  relative  units  of  a  higher  denomination. 

In  the  first  case,  fractions  are  got  rid  of 

In  the  second,  numbers  of  an  inconvenient  size  are 
avoided. 

The  first  is  transformation  descending ;  the  second, 
transformation  ascending, 

2.  In  order  to  perform  readily  such  transformations  of 
denominate  numbers  into  expressions  of  the  same  value, 
the  pupil,  instead  of  learning  by  rote  and  practising 
mechanically  certain  rules  applicable  to  the  case,  will  do 
better  to  reason  the  nature  of  the  question,  and  consider, 

I.  That  there  is,  between  the  relative  units  of  the 
same  kind  of  measure,  a  dependence,  which  is  expressed 
by  particular  names:  such  as  pounds,  shillings,  pence;  or, 
yards,  feet,  inches,  &c. 


200  LESSON   XLIX. 

II.  That  these  names  express  a  numerical  relation  or 
ratio  J  between  the  lower  and  the  higher  unit  ]  that  is,  how 
many  units  of  the  lower  denomination  it  takes  to  make  one 
of  the  higher. 

The  number  expressing  the  ratio  of  one  shilling  to  one  pound, 
is  20;  of  one  penny  to  one  shilling,  12;  to  one  pound,  240;  of 
one  foot  to  one  yard,  3 ;  of  one  inch  to  one  foot,  12,  &c. 

III.  That  an  inferior  unit  is  nothing  but  a  fraction  of 
the  superior  one ;  and  might  be  expressed,  instead  of  a 
name,  by  a  fraction  having  for  its  denominator  the  num- 
ber of  small  units  it  takes  to  make  a  large  one. 

Thus,  1  shilling  is  -^^  £>]  1  penny  is  j^  sh,,  or  ^4:^  £. 

1  ft,  is  -J-  yd. ;  1  inch  is  yV  ft.,  or  -jV  y^- 

1  qr.  is  \  cwt. ;  1  pound  is  ^i§  qr.,  or  yy ^  cwt. 
&c.  &c.  &c. 

3.  He  will,  therefore,  discover  the  identity  between 
the  change  of  unit  to  be  made  in  this  case,  and  the  trans- 
formations of  fractions. 

For  example,  let  us  suppose  that  we  want  to  change  £5  into 
shillings :  since  1  shilling  is  -^-q  of  a  pound,  it  is  transforming  the 
whole  number,  5,  into  tivciitieths,  which  gives  the  fractional 
number,  V^j  (XXX.,  8),  or,  by  using  the  word  shillhig,  instead 
of  twentieth,  into  100  shillings.  In  both  cases,  o  is  multiplied  by 
the  ratio  20  of  the  shilling  or  fraction  to  the  pound  sterling. 

If  we  wished  to  express  a  number  of  shillings;  6,  for  exam- 
ple, by  the  denomination  of  a  pound,  we  would  consider  that 
\sh.  =  £^Q'  Consequently,  6sk.  =  £,Jq.  Here  we  would  di- 
vide by  the  number  expressing  the  ratio  between  the  two. 

Likewise,  120 5^.  would  be  £^i^  =  £G. 

4.  The  identity  of  the  two  transformations,  in  frac- 
tions and  denominate  numbers,  being  well  understood,  if 
the  pupil  recollects  or  revises  what  he  has  learned  in 
fractions,  he  can  have  no  difficulty  in  performing  the 
operations,  either  in  the  case  of  augmentation  or  reduc- 
tion, of  the  number  of  units  for  the  same  value ;  which 
are  commonly  called  descending  and  ascending  reduc- 
tion. 


TRANSFORMATION   OF   DENOMINATE   NUMBERS.  201 

5.  To  transform  an  amount  of  units  from  one  denomi- 
nation into  another. 

Multiply  the  given  amount ,  by  the  value  of  one  of  its 
units,  in  terms  of  those  of  the  new  denomination. 

This  rule  applies  to  ascending  and  descending  cases, 
and  to  fractional  expressions  as  well  as  to  whole  numbers. 
For,  similar  parts  bear  to  each  other  the  same  ratio  as  the 
wholes. 

Thus,  for  example,  |  of  a  pound  is  20  times  larger  than  -J 
of  a  shilling  :  consequently, 

£}  is  20x-i  of  a  shilling,  or  \«  =  6f  shillings. 
Or,  in  decimals,       .     .     .     6.66. 

For  the  same  reason,  ^sh,  =  j£g  >^  .)q  =  ^q^^  ]   or,  in 

decimals,  0.01  66. 

In  like  manner,  j\  of  a  yard  is  3  times  larger  than  j\ 
of  a  foot ;  and  also  -^^  of  a  yard  is  3  times  larger  than 
r^-Q  of  a  foot ;  since  each  individual  unit  is  3  times  larger 
in  the  first  than  in  the  second  expression. 

Hence, 
-^^  yd.  is  3  X  {^  of  a  foot ;,  or,  |i  =  ^jj^ft. ;  or,  again,  2.\ft. 
And, 

7  i — 

hfi'  ^s  Toxl  =  To  V^-y  or?  i^  decimals,  0.023. 

By  these  considerations,  we  have  but  one  rule  for  all 
cases. 

Questions. — What  is  the  object  of  changing  the  denomination 
of  a  number  from  high  to  low?  from  low  to  high  units?  How 
is  the  dependence  of  denominate  units  expressed  ?  What  do  the 
names  indicate  ?  How  could  lower  units  be  expressed  numeri- 
cally, in  regard  to  high  ones?  Give  examples.  Is  there  any 
difference  between  the  change  of  denomination  and  transforma- 
tion of  fractions  ?  Give  examples.  What  two  cases  have  you 
to  consider?  How  do  you  proceed,  in  the  first  case?  How,  in 
the  second  ?  Does  the  same  rule  apply  also  to  fractions  ?  Show 
it  by  examples. 

EXERCISES    IN    FEDERAL    MONEY. 

Since  the  scale  of  subdivision  is  the  decimal  scale,  the 


202  LESSON  XLIX. 

removing  of  the  decimal  point  to  tlie  right,  as  was  done 
in  decimals,  will  multiply  by  10,  100,  or  1000,  and  thus 
lower  the  denomination.  On  the  contrary,  the  removing 
of  the  decimal  point  to  the  left,  will  raise  the  denomina- 
tion. 

Thus,  $15,235,  which  is  15  dollars  235  mills,  is  changed  into 
1523.5  cents — fifteen  hundred  and  23  cents  and  5  mills,  by  removal 
of  the  decimal  point  two  places  to  the  right. 

546  cents,  by  placing  a  decimal  point  two  places  to  the  left, 
will  be  changed  into  its  equivalent,  $5.46 — 5  dollars  and  46  cents. 

1.  How  many  mills  in    3  cents?  Ans.  30     mills* 

2.  «  «  10  cents?  Ajis.  100       « 

3.  "  «        in  $65  ?  .  A71S.  65,000  « 

4.  In  2  cents  and  3  mills,  how  many  mills?     Ans,  23       " 

5.  How  many  cents  in  $2  and  15  cents  ? 

6.  How  many  in  $3  and  50  cents  ? 

7.  In  $500,  how  many  dimes?  cents?  mills? 

8.  How  many  mills  in  200  dollars  37  cents  and  9  mills? 

9.  How  many  dollars,  cents,  and  mills,  in  178,854  mills? 

Ans,  178.85  4. 

10.  How  many  "  in  1,906,783  mills?     A?is.  $ 

11.  Change  19  cents ;  28cents;  45  cents,  and  9  mills,  into  mills. 

12.  "  28  mills  to  cents.  Ans.  2  cts.  8  mills. 

13.  «  209  mills  to  cents. 

14.  "  657  cents  to  dollars.  A7is.  $6.57. 

15.  "  $50  and  9  cents  to  cents.  A7is.  5,009  ce?its> 

16.  "  $75  and  21  cents  to  cents. 

17.  "  10,104  cents  to  dollars.  A?is.  $101.04. 

18.  Express  16  cents  relatively  to  dollars.  A7is.  $0.16. 

19.  «  65  mills  "  «  A?is,  $0,065, 

20.  «  5  cents  «  "  A?is, 

EXERCISES    IN   ENGLISH   CURRENCY. 

1.  Change  £17  into  shillings.  A7is.  17X20=  340  5/*. 

"  "  into  pence.                                     Atis.  4,080^, 

«  "  into  farthings.                            Atis,  lG,320/ar. 

2.  "  £256  to  shillings. 

3.  "  60  shillings  into  pence. 

4.  "  350  pence  to  farthings. 


TRANSFORMATION  OF  DENOMINATE  NUMBERS  203 

5.  Express  3  shillings  in  reference  to  a  pound.       Ans.  <£  A-, 

6.  "         6  pence  in  the  denominations  of  shillings. 

Ans*   2sh» 

7.  "         15  shillings  in  £•  Ans,  £ 

8.  "         7  pence  in  sh,  Ans.         sh, 

9.  Change  jCf  into  shillings.  Aits,  f  X  20  =  n-}sh. 

10.  "         £^  into  shillings.  A?is, 

11.  "         f  shillings  into  pence.  Ans.  Sd, 

12.  "         %sh.    '\sh.    ^sh.  into  pence.  Ans. 

13.  "         ^sh.  to  the  denomination  of  a  pound. 

Ans.  I  :  20  =  %§^. 

14.  "         f-  penny  to  the  denomination  of  a  shilling.  Ans, 

15.  "        J  penny  to  the  denomination  of  a  pound. 

^'^**   5X12X20  ~  '^TiVt). 
This  is  a  case  of  fractions  of  fractions,  -^q  of  -^2  ^^  !• 

16.  Change  £^  into  shillings,  with  decimals. 

A        4X20        o  o^ 
^;i5.  -— - —  =  8.88. 
9 

17.  "        y  shilling  to  pence,  with  decimals.         Ayis. 

18.  «         13  shillings  to  the  decimal  of  £\. 

A71S.  II  =  £0.65. 

19.  "         5  pence  to  the  decimal  of  1  shilling.       Ans. 

20.  «         £0.16  to  the  decimal  of  shillings. 

Ans.  0.16X20=  3.25^. 

21.  "         0.19  shilling  to  the  decimal  of  a  penny.       Ans. 

22.  "         0.25  shilling  to  the  decimal  of  a  pound. 

Ans.  ^^  =  £0.0125. 

23.  "         0.17  of  a  penny  to  the  decimal  of  a  shilling.     Ajis. 

0.17  r- 

of  a  pound.         Aiis.     20x12  ~  ■^0-0007083. 

24.  "        f  of  a  shilling  to  the  decimal  of  a  pound. 

^'"-  3I20  =  "-OS's- 

25.  "        -f  of  a  penny  to  the  decimal  of  a  shilling.    Ans, 
These  examples  embrace  all  the  cases  relative  to  Eng- 


204  LESSON   L. 

lish  currency.  Similar  examples  might  be  given  in  other 
measures ;  but  these  are  sufficient  to  show  the  mode  of 
operation.  The  exercises  in  compound  numbers  will 
complete  the  subject. 


LESSON  L. 
COMPOUND  NUMBERS. 

Properly  speaking,  all  numbers  are  compound,  since 
they  are  composed  of  different  orders  of  units,  connected 
together  by  a  known  numerical  relation. 

Thus,  3,756,  for  example,  is  composed  of  thousands^ 
hundreds^  tens,  and  units. 

But  the  name  of  compound  numbers  is  more  particu- 
larly reserved  for  niwibcrs  composed  of  different  denomi- 
nate units,  whose  relative  values  are  expressed  by  arbi- 
trary, names  ;  such  as  pounds,  shillings,  and  pence,  not 
connected  by  the  decimal  scale. 

Compound  numbers  are  also  susceptible  of  transforma- 
tions, or  changes  of  the  units  of  comparison,  which, 
without  altering  their  value,  render  them  more  conve- 
nient for  calculations  and  commerce. 

For  instance,  it  will  be  simpler  to  introduce  into  accounts, 
X21  65/t.,      .     .     than  its  equivalent,      .     ^^\Yl 'pence. 

In  other  cases,  for  the  purpose  of  dividing,  for  instance,  it  nnay 
be  more  convenient  to  use  5,112  pence,  tiian  its  equivalent  in 
pounds  and  shillings. 

A  moment's  consideration  will  show  that  these  trans- 
formations are  the  same  as  those  previously  explained  in 
mixed  numbers.  For,  we  may  put  a  compound  number 
under  the  form  of  a  mixed  one,  since  the  smaller  units 
are  but  fractions  of  the  larger. 

Thus,  ^£5  Ish,  may  be  written  JES^^^. 

In  that  form,  being  a  mixed  number,  the  rules  of  Les- 
son XXXVII.,  3,  may  be  applied  to  it,  and 

will  be  transformed  into  £  ^\^    =  W  ^^  ^  pound  j 


OPERATION. 

£5  Ish,  Ad,  3far, 
20 

12 

1288^. 
4 


TRANSFORMATION    OF    COMPOUND    NUMBERS.  205 

which,  by  restoring  the  name  shillings  instead  of  the 
denominator^  20,  is 

107  shillings. 

If  the  number  were  composed  of  several  descending 
units ;  as,  for  example,  in 

£o  Ish,  ^d.  Sfar. 

The  same  operation  might  be  con- 
tinued by  considering  that,  after  hav- 
ing changed  the  pounds  into  shillings, 
the  new  integer,  107 sh.  might  now, 
together  v/ith  the  next  denomina- 
tion, be  taken  as  a  second  mixed 
number,  107^^^  shillings,  and  trans- 
formed, in  its  turn,  into  twelfths  or 
pence. 

Thus,        .     107J^^  =  12X107-4-4  ^  ^^,,  shillings. 

or,  with  the  denomination  pence  instead  of  twelfths,  1,288 
pence. 

Finally,  this,  with  the  3  farthings,  would  constitute  a 
last  mixed  number, 

1,288|  pence; 

whose  transformation  into  farthings  would  complete  the 
operation,  and  give 

5,155  farthings  =  «£5  7sh.  4cZ.  Sfar. 

The  manner  of  arranging  the  operation  is  shown  above. 
It  must  have  been  noticed  that  it  is  only  a  succession  of 
transformations  into  the  consecutive  orders  forming  the 
scale,  and  which  may  be  stated  for  practice  as  follows : 

To  transform  a  compound  number  into  one  of  its  lower 
denomination. 

Beginning  with  the  higher  denomination,  transform 
successively  each  number  into  one  of  the  next  descending 
denomination  (XLIX.,  5),  adding  all  along  to  the  trans* 
formed  number  the  integers  of  the  same  denomination^ 
until  you  reach  the  denomination  required. 
18 


OPERATION. 

1264 

24 

.64 

52 

20 

I6gr. 

l^divt. 

2oz, 

206  LESSON   L. 

The  reverse  of  this  operation,  or  transformation  as* 
cending,  is  nothing  more  than  what  was  taught  in  the 
preceding  lesson.  It  is  the  same  as  the  ordinary  division 
by  which  we  change  an  improper  fraction  into  a  mixed 
number. 

Let  it  be  proposed,  for  instance,  to  change 
1264  grains  troy  weight, 
into  higher  denominations :    it 
may  be  considered  as  the  impro- 
per fraction, 

-l||4.  of  a  dwt,, 

which  becomes,  by  effecting  the  division,  the  mixed 
number 

52jf  Jtr^. ; 

or,  by  restoring  the  name  grain,  instead  of  the  denomi' 
nator,  24',  the  compound  number, 

52dwt.  16gr. 

But,  52dtDt.  being  more  than  one  ounce,  we  may  wish  to 
obtain  out  of  it  the  units  of  this  denomination.  Consi« 
dering  that 

52dwt.  =  II  of  an  ounce, 

and  that  the  improper  fraction, 

f  §0^.  =  2|§02;. ; 

or,  as  a  compound  number, 

2oz.  12dict., 

we  get,  finally,         l,26%r.  =  2oz.  12divt.  16gr, 

The  manner  of  arranging  the  operation  is  shown  above, 
and  it  appears  that  the  last  quotient  and  the  successive 
remainders  compose  the  result.  Here,  again,  we  see 
only  a  repetition  of  known  operations ;  and  we  have 
hardly  occasion  for  the  following  rule  : 

To  transform  a  large  number  of  a  lower  denomination 
into  a  compound  number  of  its  higher  denominations. 

1 .  Divide  the  number  by  the  ratio  of  its  units  to  thosf 
of  the  next  higher  denomination. 


TRANSFORMATION   OF    COMPOUND   NUMBERS.  207 

2.  Then  divide  the  quotient  in  the  same  icay,  and  each 
svccessive  quotient  until  you  reach  the  higher  denomina- 
tion required, 

3.  The  answer  will  be  the  last  quotient,  with  all  the 
remainders  in  their  order » 

Questiojis. — What  is  a  compound  number  ?  May  a  compound 
number  be  expressed  as  a  mixed  number?  How?  Give  exam- 
ples. In  what  does  the  change  of  units,  in  compound  and  in 
mixed  numbers,  differ  ?  Give  the  rule.  What  name  is  given  to 
the  transformation  from  a  higher  to  a  lower  unit?  From  a  lower 
to  a  higher  ?  What  simple  operation  is  the  last  ?  Give  an  ex- 
ample. What  is  each  quotient ;  what,  each  remainder?  Is  the 
result  a  compound  or  mixed  number  ?     Repeat  the  rule. 

EXERCISES. 

1.  In  £91  llsJi.  SJ^^.,  how  many  farthings?        Ans,  81,902far, 

2.  In  19.sh.  8|^.,  "  "  A7is,  Mlfar, 

3.  Reduce  38-5^.  A^d.  to  halfpence.  A7is.  921  halfpence, 

4.  How  many  pounds,  &c.,  in  3,175/ar.  ?       Ans.  £3  6sk.  l^d, 

5.  "  «  inl,206J.?  A?is.  £5  Osh.  Qd, 

6.  How  many  shillings  in  9,1 52d.?  Ans.  S12sk.  8d, 

7.  How  many  pounds  in  143</.  ?  A7is.  £0  lls/i.  lid, 

8.  In  £121  O.sh.  9ld.,  how  many  halfpence?  A^is.  58,099. 

TROY   WEIGHT. 

1.  In  151b.,  how  many  grains?  Ans.  86,400o'r. 

2.  How  many  ounces  in  5,749^/2^;^.?  A7is.  281oz.  9dwt, 

3.  Reduce  llo^.  ISdivt.  13gr.  to  grains.  A7is.  5,605gr. 

4.  How  many  grains  in  15  spoons,  each  weighing  Qdwt.  15gr.  ? 

A71S.  2,385gr, 

5.  Reduce  47/i.  lOoz.  15dwt.  to  d^vt,  A7is.  ll,4:95dwt, 

6.  In  lQ,113gr.,  how  many  pounds? 

A71S.  13lb.  ooz.  18dwt,  21gr, 

AVOIRDUPOIS. 

1.  In  19T.  Ucwt.  2qr.  19/3.  llo^.  13dr.,  how  many  drams? 

Ans.  ll,316,157^r. 

2.  How  many  cwt.  in  9,563/3.?  Aiis.  85cwt.  Iqr.  1516, 


208 


3.  In  13 civt.  211  h.,  how  rndLuy -pounds  1  Ans.  1,477/5. 

4.  In  Sr.  25/5.,        «  «  -4?z5.  6,745/5. 

5.  Reduce  Icwt.  3qr.  11/5,  to  ounces.  A?is,  14,064/5. 

6.  Change  8,994,384  ounces  to  tons,  &c. 

A71S.  250 T.  19cwt.  Oqr.  21/5. 

apothecaries'  weight. 

1.  Reduce  9Ib  2g  3^  1^  Igr.  to  grains.  Ans.  53,007^r. 

2.  «  16Ib  6^  to  ounces.  Ajis.  192g  63. 

3.  «  15B5  to  grains.  A71S.  S6,400^r. 

4.  «  2,9259  to  pounds,  &c.  A7is.  10B5  Ig  7^. 

5.  "  2,107^r.  to  ounces,  &c.  Aiis.  4g  3^  OQ  7^r. 

6.  "  58,478o-r.  to  pounds.  A71S.  lOB)  1"^  63  IQ  IS^-r. 

WINE    MEASURE. 

1.  In  25  tuns,  how  many  pints  ?  Aiis.  50,400. 

2.  How  many  hogsheads  in  4,935  quarts  ? 

A?is.  19khd.  36gal.  3qt. 

3.  In  3hhd.  13gal.  2qt.,  how  many  half  pints?  A7is.  3,240. 

4.  In  52  gallons,  how  many  gills  ?  A71S.  1,664. 

5.  Reduce  2  hogsheads  50  gallons  to  pints.  A7is.  1,408. 

6.  Reduce  5  hogsheads  to  pints.  Atis.  2,520. 

ALE    OR   BEER   MEASURE. 

1.  Change  5  hogsheads  to  pints.  A7is.  2,160, 

2.  In  6  hogsheads  2  quarts,  how  many  pints  ?  ^725.  2,596. 

3.  In  665  pints,  how  many  hogsheads  ? 

Ans,  Ihhd.  29gal.  Qqt,  Ipt, 

4.  Change  375«r.  6gal.  3qt.  to  pints.  Ans.  10,710. 

DRY   MEASURE. 

1.  Change  72  bushels  into  pints.  A71S.  4,608. 

2.  "        15  bushels  Specks  2  quarts  into  pints.      Ans.  1,012. 

3.  In  16  chaldrons  15  bushels,  how  many  pecks?     Ans.  2,364. 

4.  Change  1,024  pints  into  bushels.  A71S.  16. 

5.  "        5,328  pints  into  chaldrons.         A71S.  2ch.  llhi.  Ij)^. 

6.  In  6,225  bushels,  how  many  barrels  of  corn?        Ans. 


TRANSFORMATION    OF    COMPOUND    NUMBERS.  209 

LONG    MEASURE. 

1.  In  57  miles  2  furlongs,  how  many  poles  ?  A?is.  18,320. 

2.  Plow  many  furlongs  in  19,753  yards?       A?is.  Hdfur.  113yd. 

3.  In  590,057  inches,  how  many  leagues? 

Aus.  3lea.  2ficr,  llOyd.  1ft,  5ut. 

4.  In  38,396  rods,  how  many  miles?     A7is.  1197nl.  Ifur.  SQrd. 

5.  In  75  degrees,  how  many  miles?  Ans.  5,180. 

6.  In  2,085  miles,  how  many  degrees?  Ajis. 

7.  Reduce  4  yards  5  feet  11  inches  to  inches.  A7is.  215m. 

8.  In  4,652  yards,  how  many  miles?  Ans.  2mL  1,132yd. 

CLOTH   MEASUPvE. 

1.  In  158  yards,  how  many  nails  ?  A^is.  2,528. 

2.  How  many  ells  English  in  5,932  nails  ?         Ans.  2d6E.  3qr. 

3.  In  29  pieces  of  holland,  each  of  36  ells  Flemish,  how  many 
yards?  A71S.  783. 

4.  Change  42  English  ells  3  quarters  to  quarters.        Ajis.  213. 

5.  In  17  yards  2  quarters  2  nails,  how  many  nails  ?     A7is.  282. 
6..  In  65  ells  French,  how  many  yards?  A?is.  91yd.  2qr, 

LAND    OR    SQUARE    MEASURE. 

1.  In  41  acres  2  roods  14  perches,  how  many  rods? 

Ans.  6,654P. 

2.  How  many  square  rods  in  7,752  square  feet? 

A71S.  28P.  129Sq.  ft. 

3.  In  5,972  perches,  how  many  acres  ?    .    Ans.  31  A.  IR.  12P. 

4.  Change  12  square  yards  to  square  inches.  Ans.  15,552. 

5.  Change  6,480  square  inches  into  square  yards.  A7is.  5. 

6.  In  25^.  3R.  12P.,  how  many  square  poles?         A7i3.  4,132. 

SOLID    OR    CUBIC    MEASURE. 

1.  In  a  pile  of  wood  96ft.  long,  5ft.  high,  Aft.  wide,  how  many 
cords?  Ans.  15. 

2.  In  82  tons  of  round  timber,  how  many  inches  ? 

A71S.  5,667,840. 

3.  How  much  wood  in  a  load  6ft.  long,  Aft.  high,  2|/jf.  wide? 

A71S.  41  cord,  or  3£  cord  feet, 

18*  o         32       .       *       y 


210  LESSON   LI. 

4.  In  6  cubic  yards,  how  many  inches?  Ans.  279,936. 

5.  In  64,000  cubic  inches,  how  many  cubic  yards  ? 

Ans.  lyd.  IQft.  QUn, 

6.  How  many  cubic  inches  in  a  cord  of  wood?     A7is.  221,184. 

TIME. 

1.  How  many  hours  in  57  years?  Ans.  499,662. 

2.  In  57,953  hours,  how  many  weeks  ?     Ans.  Miwk.  6da.  llhr, 

3.  How  many  days  from  19th  March  to  23d  September  ?     A7is, 

4.  How  many  days  from  19th  November,  '43,  to  14th  May,  '44  ? 

Ans. 

5.  How  many  seconds  in  a  year?  Ans.  31,557,600. 

6.  How  many  in  a  day  ?  Ans.  86,400. 

7.  If  a  man  can  count  100  per  minute,  how  long  will  it  take 
him  to  count  1,000,000,  at  the  rate  of  10  hours  per  day  ?   Ans. 

CIRCULAR   MEASURE. 

1.  Reduce  6''  9'  to  minutes.  Ans.  369'. 

2.  "         9°  40'  44"  to  seconds.  Ans.  34,844". 

3.  How  many  minutes  in  the  circle  ?  Ans. 

4.  How  many  seconds  ?  Ans. 

5.  How  many  degrees  in  3,315"?  Ans.  0°  55'  15". 

6.  How  many  signs  in  2,021'?  Ans.  IS.  3^  41'. 

LESSON  LI. 

1.  Compound  numbers  are  sometimes  subjected  to  a 
kind  of  transformation,  which  it  may  be  proper  to  notice ; 
though,  if  properly  investigated,  it  also  falls  under  some 
of  the  rules  previously  explained ;  and  a  little  considera- 
tion would  enable  the  pupil  to  perform  the  operations  by 
means  of  the  principles  he  is  already  acquainted  with. 

These  transformations  consist : 

1st.  In  expressing  a  denominate  fraction  of  a  higher 
unit  in  integers  of  its  lower  denominations. 

2d.  In  the  inverse  operation,  which  consists  in  express- 
ing a  compound  number  as  a  denominate  fraction  of  a 
hiojher  denomination. 


TRANSFORMATION   OF    COMPOUND   NUMBERS.  211 

CASE    I. 

2.  Let  us  take,  as  an  example  of  ths  first  case,  the  de- 
nominate fraction, 

•  ^1, 

to  be  changed  into  integers  of  a  lower  denomination  :  by 
turning  to  Lesson  XLIX.,  the  pupil  will  recognise  here 
a  case  of  descending  transformation  of  denominate  frac- 
tions ;  and,  by  a  known  rule,  he  will  get 

je|-  =  1005^.  =  l^sh. 

But  the  fraction,  -^-  or  |  sh.^  in  its  turn,  may  be  changed 
into  pence, 

^  3 

Hence,  the  complete  transformation  is 

£i  =  I65A.  8^. 

The  operation  is  evidently  the  same  as  would  take 
place  in  the  extension  of  a  division  to  lower  units.  If, 
for  instance,  the  remainder  of  a  division  by  6  had  been 
j65,  and  we  wished  to  get  the  part  {-  of  the  quotient  in 
smaller  units,  the  operation  would  be  as  follows : 

1st.  Multiply  £b  by  20,  to 
change  it  into  shillings. 

2d.  Divide  the  product,  100, 
by  6,  and  the  quotient  is  165^., 
with  a  remainder,  4. 

3d.  Multiply  this  remainder 
by  12,  to  lower  to  pence. 

4th.  Divide  the  product,  48, 
by   6,  and   the    quotient,    ScZ., 
completes  the  answer,  which  is  composed  of  the  succes- 
sive  quotients. 

If  there  was,  again,  a  remainder,  you  might  proceed 
in  the  same  way,  to  farthings.  This  is  analogous  to  what 
was  done  in  decimals ;  and  it  is  hardly  necessary  to  state 
that,  in  order  to  change  a  denominate  fraction  to  inte- 
gers of  a  lower  denomination, 


OPERATION. 


5         I    6 
20 


16^^.  8^. 


1005/i. 

4 
12 

48^.  " 


212 


LESSON   LI. 


OPERATION. 

0.7  bushels. 
4 

2.8  pecks. 
2 

1.6  gallons. 
4 


Transform  the  numerator  of  the  fraction  into  lower 
tinits  and  divide.  If  there  is  a  fractional  part  in  the 
quotient,  transform  it  in  the  same  manner  until  you  reach 
the  lower  denomination  required. 

The  operation  would  be  the  same,  though  rather  easier, 
with  decimal  fractions. 

Let  us  take,  as  an  example, 

0.7  of  a  bushel. 

By  the  principles  of  Lesson 
XLIX.,  you  change  this  into 
pecks  by  multiplying  by  4,  and 
thus  get 

2.8  pk,,  or  2  pics,  and  0.8  of  a  pJc, 

Now,  in  its  turn,  the  fraction-  2.4  quarts. 

al  part,  2 

0.8  multiplied  by  2,  is  1,6  gal. 

Then  the  fractional  part,  .6,  multiplied  by  4,  gives 

2.4  quarts; 

the  fractional  part  of  which  contains  no  integer  of  pints. 
Therefore,  the  answer  is,  by  collecting  all  the  integers  of 
the  successive  products, 

O.lbush,  =  2p1cs.  Igal,  2,^qt. 

What  renders  here  the  operation  easier  is,  that  the  use 
of  the  decimal  point  dispenses  with  divisions. 

N.  B. — Be  careful  to  multiply  only  the  decimal  part  of  each 
product,  which  is  the  only  one  to  be  transformed. 

CASE    II. 

3.  In  the  transformation  of  a  compound  number  into 
a  fraction  of  a  higher  denomination,  the  pupil  will  readily 
recognise  an  addition  of  fractions. 

Let  it  be  proposed,  for  example,  to  convert 
9o2J.  6dwt.  \6gr. 


TRANSFORMATION   OF    COMPOUND    NUMBERS.  213 

into  the  fraction  of  a  pound  troy.  It  is  clear  that,  if  instead 
of  employing  names,  we  express  the  successive  numbers 
by  their  fractional  values,  relative  to  the  pound,  the  com- 
pound number  will  become 

(t2  +  2V  of  t\  +  -2I  of  2V  of  tV)  of  a  pound, 
or,        A-  +      ^      + 15_  =:  m. 

So  that  the  operation  is  nothing  else  than  an  addition 
of  fractions,  and  may  easily  be  performed  as  such. 
The  usual  method,  however,  is  as  follows : 

Reduce  the  compound  number 
to  its  lowest  denomination,  and 
set  the  result,  as  a  numerator, 
over  the  number  ivhich  expresses 
the  ratio  of  the  last  units  to  the 
higher  denomination  required. 

Reduce  the  fraction  to  its 
lowest  terms. 

In  the  example  we  have  chosen,  the  compound  num- 
ber is  changed  to  4,480^r.  j  and  Igr.  being 

24X20X12  ^^  ^  ^^-  5   °^J  i>TVo^^-J 

the  answer  is  fyl^?^.,  which  can  be  transformed  mio  ^Ib, 

4.  The  transformation  of  a  compound  number  into  a 
decimal  fraction  of  a  higher  unit,  might  be  made  in  two 
different  ways : 

I.  Either  find  the  vulgar  fraction  equivalent  to  the 
compound  number,  and  change  it  into  a  decimal  fraction  ; 

II.  Or,  change  successively  each  part  of  the  compound 
number  into  a  decimal  fraction  of  the  denomination  just 
above  it,  until  you  reach  the  upper  one. 

This  second  method  will  generally  be  found  the  short- 
est. Let  us  take  the  same  example  as  above,  and  consider 
each  denominate  part  successively. 

The  third,     .         16^r.  =  l\dwt.  =-.  O.'Mwt. 


OPERATION. 

9or.  Qdtvt.  IQgr, 
20 

ISQdwt, 
24 

4480^r.=  |f|4=i/5. 


214 


LESSON    LI. 


This,  together  with  Gdivt.,  is  6,^6dwt.  —  ^f§oz.  =z 
0.3'3oz.',  which,  added  to  9oz,,  gives ^-f^lb,  =  0^.7778^6. 

The  operation  is  given  here  in  all  its  details,  but  may 
be  performed  almost  at  a  glance,  with  but  little  practice. 

Questions. — How  do  you  change  a  denominate  fraction  into  a 
compound  number  of  lower  unfits?  What  simple  operation  is 
it  ?  How  is  the  same  done  in  decimals  ?  Give  examples.  How 
do  you  transform  compound  numbers  into  a  denominate  fraction 
of  a  higher  denomination  ?  What  simple  operation  is  it  ?  What 
is  the  usual  way  ?  How  is  the  same  transformation  made  in  de- 
cimals? 

EXERCISES   IN  TRANSFORMATION  OF  DENOMINATE  FRACTIONS, 
BOTH   VULGAR   AND   DECIMAL. 

N.  B. — Look  to  the  rules  and  examples  in  Sterling  Money. 

1.  What  part  of  sl  dwt.  is  -j^  of  a  lb.  troy  ?  Ajis.  j» 

2.  Change  j-^cwt.  to  the  fraction  of  a  lb.  Ans.  y* 

3.  Change  f  of  a  i?  into  smaller  units. 

Ans.  f  X^Osk.=  13^sh. 
And  because         .        .        ^sh.  =  4d.         .         .        ==13^^.46?. 

4.  Change  f  of  a  shilling  to  smaller  denominations. 

Ans.  I  X  12d.  =  4id. 

4X4 
And  because         .         ^d.  =  — 5~~/«'"-  •  =  4^?.  3^far, 

5.  Express  -f-  lb.  troy  in  smaller  integers.         Ans.  loz.  4dwt, 

Ans.  Qfur.  16P. 

7.  Reduce  2qr.  o^na.  to  the  fraction  of  an  ell  English. 

Ans.  \E.  E. 

8.  Reduce  4sh.  Q\d.  to  the  fraction  of  a  £.  Ans.  £\%^* 

9.  Express  Ush.  5ld.  in  decimals  of  a  £.     Ans.  £0.1229l'6 

10.  «        I5sh.  in  decimals  of  a  £.  Ans.  £0.15. 

11.  "         3qr.  18lb.  in  decimals  of  a  cwt.     Ans.  0.910714  + 

12.  "         2qr.  2na.  in  decimals  of  a  i/d.  Ans.  0.625. 

13.  "         14gal.  3qt.  of  wine  in  decimals  of  a  hogshead. 

Ans.  0.2341  + 

14.  "         \3sh.  9|^.  in  decimals  of  a  £.  Ans.  0.691. 

15.  What  is  the  value  of  0.9  shilling?  Ans.  lOfrf, 


COMPOUND   ADDITION.  215 

16.  Change  0.592  of  a  cwt.  into  a  compound  number. 

Ans.  2qr.  lOlb.  4.0Z.  13d?/-. -f- 

17.  Express  0.25S  tun  of  wine  in  lower  integers. 

Ans.  Ihhd.  ^gaL-^" 

18.  «         0.12785  of  a  year.      A7is.  46da.  15hr.  57^.  57" -|- 

19.  "        2ft.  3in.  in  the  fraction  of  a  yard.  Ans.  ^yd. 

20.  What  is  |-  of  a  hogshead  ?  Ans.  52gal.  2qt. 

21.  What  is  I  of  a  guinea?  Ans.  IQsh.  9|^' 

22.  What  part  of  a  mile  is  9ft.  9m.  ?  Afis.  T^j^-Qml. 

23.  What  part  of  a  day  is  Ahr.  3m.  ?  Ans.  -^e^da, 

24.  Change  lib.  4oz.  12dtfft.  IGgr.  to  the  decimal  of  a  pound. 

Ans.  1.3861/3. 

25.  Express  6,600/jf.  in  decimals  of  a  mile.  Ans.  1.25ml, 

26.  What  is  in  integers  0.625  of  a  gallon?  Ajis.  2qt.  Ipt, 

27.  What  is  .025  of  an  acre  ?  Ajis.  4P. 

28.  What  is  .  125  of  a  year,  leaving  out  the  6  hours  ? 

Ans.  4:5da.  15hr. 

29-  What  is  2.16  miles?  Ans.  2^1.  51rd.  1yd.  Qtft.  Sfin. 

30.  What  is  0.1125  of  a  ton?  A^is,  2cwt.  Iqr, 

31.  What  decimal  fractions  of  Icwt,  is  1/3.?     Ans,  .00892857. 

32.  What  decimal  fraction  of  a  pound  is  1  farthing  ? 

Ajis.  .001041^ 


LESSON  LII. 
COMPOUND  ADDITION. 

RULE. 

1.  I.  Set  down  the  units  of  the  same  denomination 
under  each  other. 

n.  Take  the  sum  of  the  column  of  the  lowest  denomina- 
tion ;  find  by  division  how  many  units  of  the  next  higher 
order  there  may  he  in  this  sum  ;  set  down  the  remainder 
under  the  column  just  added  up,  and  carry  the  units  of 
the  quotient  to  he  added  with  the  next  column, 

in.  Proceed,  in  the  same  way  with  all  the  columns. 


1?16 


LESSON   LII. 


FIRST   EXAMPLE. 

2.  For  a  first  illustration  of  this  rule, 
let  us  take  the  annexed  addition  in  avoir- 
dupois weight. 

The  addition  of  the  pounds  gives  40, 
that  is  1  quarter  and  12  pounds;  I  set 
down  12  and  carry  1  quarter. 


OPERATION. 

cwt.  qr.  lb. 

76  3  14 

37  2  15 

14  1  11 

128      3      12 


Now  1  quarter  added  to  the  others  makes  7  quarters, 
which  are  1  cwt.  and  3  quarters;  I  set  down  3  and  carry  1. 
Finally,  adding  1  together  with  the  cwt.  gives  128  cwt. 


£ 

sh. 
83 

d. 

765 

19 

7 

1,279 

17 

6 

915 

13 

11 

2,594 

19 

8 

Sum  total, 

689 

8 

6 

£6,145 

195^ 

.2d, 

SECOND  EXAMPLE. 

Let  it  be  proposed  to  add  the  following  numbers : 

Adding,  first,  the  column 
of  pence,  I  get  38  pence, 
which  contain  3  shillings 
and  2  pence. 

I  set  down  2  under  the 
column  of  pence,  and  carry 
the  3  shillings  to  the  column 
of  units  of  shillings. 

Adding  up  the  first  column  in  shillings,  I  find  39 ;  I 
set  down  9  and  carry  3  tens,  to  be  added  with  the  second 
column  of  shillings. 

I  now  add  the  tens  of  shillings  and  get  7 ;  and,  because 
it  takes  2  tens  of  shillings  to  make  one  pound,  I  take  one 
half  of  7,  which  is  3,  with  a  remainder,  1.  I  set  down 
this  remainder  and  carry  the  £3  to  the  column  of  pounds. 

Then  I  complete  the  addition,  and  the  result  is  as  above. 

These  two  examples  will  suffice  to  explain  the  rule  for 
the  addition  of  compound  numbers. 

3.  This  addition  is  proved  like  simple  addition,  by  re- 
versing the  operation. 

Questions. — What  is  compound  addition?  How  do  you  set 
down  the  numbers  ?  Where  do  you  begin  to  add  ?  What  do  you 
do  with  each  successive  sum  ?  What  do  you  set  down  ?  What 
do  you  carry  ?    How  do  you  prove  addition  ? 


COMPOUND    ADDITION. 


217 


EXERCISES. FEDERAL   MONEY. 


$75  3c.  ein.  +  $25 1m. 
-+-$25615f.4-$618w. 
-f  278w.  +  96c. 

(2.) 
$2095^.4-65w.+ 
273ci5.4-$6  3c. + 
$44  _j_  $9,729  68  c. 

(3.) 

$719  2c.  +  6,921w.4. 
3,456c.  +  $0.15+$900 
-|-67c^5.4-$2;966. 

] 

ENGLISH    CURRENCY. 

£     sh,     d. 

£      sh,      d. 

£ 

sh.        d. 

1       3       5 
1       0       9 
0       7     llf 
20     13       9 
0       1       3f 

149     14       6f 
387     19       8| 
456     15       7i 
622       0       8 
734       9       9| 

28 
605 
100 

77 
0 

9       lOi 

3  7f 

4  0 
2         8i 

13       llf 

23 


2i 


lb,  oz,  dwt,  gr, 

6       2     11     19 

64       9     17     22 

0       6       0     17 

100       7     19     23 

172       2     10       9" 


TROY   WEIGHT, 

oz.    dwt,   gr, 
37       9       3 
9       5       3 
16     21 

7       8 


3 

17 


9       0 
0     19 


Ih.    oz.  dwt.  gr, 
83     11     15     22 


16 
10 
32 
44 
18 


19  20 

0  9 

8  10 

17  14 

16  7 


AVOIRDUPOIS    WEIGHT. 


t07is.  cwt,  qr.  II,    oz, 
2       2       1     17     10 


12  10 
0  2 
0       0 


2 

IS 


2 

3 

11 


14     14       3     19     10 


cwt,     qr,    lb. 


624 
207 
159 
211 
0 


19 

25 

27 

0 


civt.  qr.    lb.   oz.   dr. 

91     2     15     14       1 

21 


25     3 
19     0 


33 
10 


0 
6       2 
16     15 


4 


15 
2 
10     11 


ffi 

2 

3  9 

gr. 

11 

6 

2   1 

15 

25 

11 

7  2 

19 

44 

0 

5  0 

11 

36 

9 

2  2 

18 

3 

0 

6     0 

17 

121 

5 

1     0 

19 

0 

APOTHECARIES'  WEIGHT. 

ft  I  Z  'b  gr. 

25  7  7  2       0 
9  10  0  1  19 

26  2  1  2  6 
34  11  6  0  12 
16  7  5  1  17 


ft  ^  Z  b  gr- 

34  0  2     1     6 

15  7  7     2     9 

165  8  3     1  16 

22  6  6     0     0 

49  5  5     2  18 


218 


LESSON  LII. 


CLOTH   MEASURE. 

(1.)  (2.)  (3.) 


yd, 
150 

qr, 
3 

2 

29 

qr,     na, 
2         1 

yd, 
210 

qr. 
3 

na, 
3 

219 

1 

0 

36 

1    3 

25 

1 

1 

501 

2 

3 

456 

0    2 

97 

2 

3 

61 

3 

2 

35 

1    3 

0 

1 

2 

0 

1 

1 

9 

0    0 

37 

2 

3 

934 

0 

0 

LONG 

MEASURE. 

16 

ml,  fur 
2   7 

.P. 

25 

yd. 

3 

1 

mL  fur,  P, 
29  3  14 

yd, 

127 

ft. 
2 

5 

25 

1   6 

17 

2 

2 

19  6  29 

12 

2 

9 

10 

0   1 

0 

1 

0 

5  4  20 

0 

2 

0 

44 

2   5 

39 

5 

1 

9  1  37 

3 

1 

11 

0 

1   4 

31 

4 

2 

7  0   3 
4  5   9 

213 
25 

2 
0 

10 
0 

98 

0   1 

35 

0 

01 

• 

SQUARE    MEASURE. 
M,    A.    R,   P.  lyr.    A.    R,    P.     Sq.yd.  Sq.ft.  Sq, in. 


6  500  2 

36 

7  320  1 

13 

19    5 

141 

19   21  3 

15 

10   10  0 

10 

25    7 

126 

20  375  2 

17 

75  415  2 

31 

9    8 

117] 

11    6  0 

19 

6    3  0 

15 

21    4 

•19 

0  615  1 

39 

19   18  3 

36 

16    3 

100 

15  575  1 

17 

7    0 

109 

58  239  3 

6 

CUBIC  MEASURE. 

Cyd.  C.ft. 

C.in, 

Cords.  C.ft. 

C.in. 

Cyd.  C.ft. 

C.in. 

51    15 

1678 

15   127 

1600 

15   6 

1500 

76    5 

1000 

2    61 

1521 

21  21 

0 

33    20 

567 

4    72 

304 

697  14 

212 

1    19 

1062 

21    78 

1411 

4322  19 

699 

3    0 

9 

44   106 

319 

96   0 

247 

15    26 

1708 

16   123 

0 

149  26 

1461 

182  7         840 

WINE    MEASURE. 
T.pi.hhd.  bar. gal.  qt.        T.  pi.  hhd.  bar. gal.  qt.     hhd.  gal.  qt.  pt. 


21  1 

1 

1 

30 

3 

6 

1 

] 

31 

2 

17 

60 

3 

1 

72  0 

1 

0 

25 

2 

15 

0 

0 

26 

3 

2 

51 

2 

0 

21  1 

1 

0 

31 

1 

26 

0 

1 

15 

1 

5 

25 

1 

1 

4  1 

0 

1 

7 

2 

31 

0 

1 

3 

0 

9 

17 

0 

1 

3  1 

0 

0 

16 

3 

71 

9.1 

1 
1 

0 
0 

18 
9 

3 
3 

7 
16 

44 
19 

2 
3 

1 

1 

124  0 

1 

! 

17 

1 

COMPOUND   SUBTRACTION.  219 


DR^ 

'  MEASURE. 

(1.) 

(2.) 

(3. 

th. 

^M.  ^Z'.  qt,  pt» 

eh. 

hu,  pk.  qt»  pt, 

lu.pk.gal.qt.pt. 

16 

26  3  3  1 

19 

30  2  6  1 

6  2  12  1 

10 

35  2  7  1 

21 

18  3  1  0 

21  3  1  3  0 

0 

16  1  6  0 

36 

17  0  7  1 

17  0  0  0  0 

2 

0  12  1 

0 

14  1  0  1 

18  1  1  2  1 

4 

19  0  4  0 

7 

32  3  5  0 
0  3  4  0 

26  3  0  3  1 
0  3  13] 

34~ 

26  1  7  1 

9 

TIME. 

yr. 

mo,  da.  hr,  '     " 

yr. 

mo.  da,  hr.  ' 

'/ 

mo,  da.  hr.  ■*    " 

4 

10  15  7  9  17 

115 

0  0  0  0 

0 

1  4  6  2  17 

7 

9  29  16  51  49 

17 

7  17  17  41 

19 

26  7  3  40  19 

10 

6  13  9  0  0 

91 

5  6  10 

48 

6  19  17  14  45 

15 

7  22  14  56  40 

29 

11  28  22  52  ; 

24 

2  2  2  10 

6 

0  1  2  24  59 

10 
0 

3  6  17  25 
0  14  1 

26 

41  17  3  18  53 
21  21  21  21  21 

iT 

10  22  2  22  45 

7 

CIRCULAR   MEASURE. 


6  15 

59 

21   3   8 

17  32  57 

9  14 

41 

1  16  19 

6  17   8 

10  32 

26 

31   3   2 

3   9   1 

14  41 

32 

6   8  59 

4  50  50 

2   0 

1 

7  11  30 

7   9  10 

7  42 

11 

9  39  17 

8  19  34 

50  26 

50 

LESSON  LIIL 
COMPOUND  SUBTRACTION. 

RULE. 

I.  T.  Place  the  less  number  below  the  greater^  so  thcd 
like  units  may  be  exactly  under  each  other, 

II.  Begin  with  the  lowest  order,  and  subtract  each 
figure  from  the  one  above  it,  if  possible, 

III.  But^  if  a  number  of  any  order  is  larger  than  that 
above  it,  make  the  subtraction  possible  by  adding  to  the 


220 


LESSON   LIII. 


upper  number  as  many  units  as  are  equal  to  one  of  the 
next  higher  order;  subtract  and  carry  1  to  the  loicer 
number  of  the  next  higher  order ;  go  on,  in  the  mme 
way,  to  the  end. 

2.  It  will  be  observed  that  this  is  analogous  to  the  sub- 
traction of  simple  numbers,  and  exactly  the  same  as  the 
rule  for  subtracting  mixed  numbers,  as  it  should  be;  since 
a  compound  number  is  in  fact  a  mixed  number,  and  all 
the  operations  with  it  should  be  performed  in  the  same 
way. 

To  illustrate  this  rule  by  an  example,  let  us  suppose 
that  we  have,  £      sh,     d,  far. 

From         .         .         .        .        25       0       5     2 
to  subtract      .         •         .         .         17     11     10     1 


£7       85^.  Id.  Ifar. 

In  this  example,  we  subtract  1  farthing  from  2,  and  set 
down  the  remainder,  1. 

Then,  as  we  cannot  take  lOd,  from  5,  we  render  the 
subtraction  possible  by  adding  to  5  the  value  of  1  shil- 
ling; that  is,  12J.,  and  say,  10  from  17,  leaves  7, 

Now,  having  added  the  value  of  1  shilling  to  the  upper 
number,  we  must  add  the  same  to  the  lower  one,  in 
order  that  the  difference  may  not  be  changed.  We  there- 
fore say,  1  shilling  carried  and  11  are  12;  12  from  0  we 
cannot.  Here,  again,  to  render  the  subtraction  possible, 
we  add  to  the  shillings  the  value  of  1  pound ;  that  is,  20 
shillings,  and  say,  12  from  20  leaves  8,  which  we  set 
down. 

Finally,  the  pound,  added  as  20  shillings  to  the  upper 
number,  must  also  be  added  to  the  lower  one ;  and  we 
conclude  by  saying,  1  carried  and  £17  are  18,  which, 
taken  from  25,  leaves  £7. 

PROOF. 

3.  Compound  subtraction  is  proved  as  in  simple  num- 
bers, by  adding  the  smaller  number  and  remainder. 


COMPOUND  SUBTRACTION.  221 

EXAJfPLE.  OPERATION. 

mo.  day    h.    m.    sec. 

Let  it  be  proposed  from  6     16     13     27     19 

to  subtract  1     22     16     41     37 

42  sec.  and  37  are  one  min.  and  19  sec. 

46  min.  and  41  are  one  hour  and  27  min.  4     23     20     45     42 

21  h.  and  16  are  one  day  and  13  h.        Proof  6     16     13     27     19 
24  days  and  22  are  one  mo.  and  16  days. 
Finally,  5  mo.  and  1  are  6  mo.  which  proves  the  operation. 

Questions. — How  are  numbers  set  down  in  subtraction  ?  Where 
do  you  begin?  If  the  lower  number  of  any  denomination  is 
larger  than  the  upper  one,  how  do  you  subtract?  Why?  Give 
an  example.  What  do  you  add  ?  What  do  you  carry  ?  Repeat 
the  rule.     How  do  you  prove  subtraction? 


EXERCISES,  WITH  PROOFS. FEDERAL  MONEY. 

1.  From  215  dollars  6  cents,  subtract  119  dollars  55  mills. 

2.  From  16  dollars  and  20  mills,  subtract  7  dollars  25  cents. 

3.  From  6,728  cents,  subtract  50,739  mills. 

ENGLISH    CURRENCY. 

1.  From  £1  2sh.  Id.,  take  3sk,  Ad,  Ans.  \Ssh,  9d. 

2.  From  £5  2sh.  Id.,  take  9sk.  6d,  Ans.  £4  12sh,  Id. 

3.  From  £8,  take  205/^.  and  3far.  Ans.  £6  19^^.  llj^. 

TROY   WEIGHT. 

1.  From  lib.  3oz.  llgr.,  take  3lb.  loz.  5dtvt.  \9gr, 

2.  From  Alh.  9oz.  Idwt.  13dr.,  take  3lb.  8oz.  iddwt.  12gr. 

3.  From  73/5.,  take  26/5.  loz.  2dwt.  16^r. 

AVOIRDUPOIS    WEIGHT. 

1.  From  32r.  \()cwt.  Igr.  27/5.,  take  142".  19cwt.  3qr.  16/5. 

2.  From  3cwt,  2qr.  lib.  6oz.  15dr.,  take  2cti^.  Sgr.  25/5.  lozm 
Udr. 

S.  From  16^.  16/5.,  take  5cwt.  15oz. 

N.  B. — Let  these  also  be  done  when  the  hundred  weight  is 
only  100  pounds. 

19* 


222  LESSON   LIII. 

APOTHECARIES'    WEIGHT. 

1.  From  lOffi  12gr.,  take  m  65  IQ  I3gr. 

2.  From  1121b  73  SQ  19^r.,  take  97ft  11^^  63  29  1^. 

3.  From  163),  take  9Ib  lOg  7^  ig  IS^-r. 

CLOTH   MEASURE. 

1.  From  dSiyd.,  take  150yd.  3qr.  2?ia, 

2.  From  69E.FL  2qr.  hia.,  take  50E.FL  3na, 

3.  From  51-E.  Fr,  5gr.  2?ia.y  take  51y^.  3qr.  3?ia. 

LONG    MEASURE. 

1.  From  IGml.  bfur,  3\rd.  Ayd.,  take  5mL  5yd. 

2.  From  35rd.  2yd.  2ft.  2m.,  take  IQrd.  3yd.  2ft.  \Oln, 

3.  From  17^.,  take  39r^.  ^yd.  \ft,  lliw. 

SQUARE   MEASURE. 

1.  From  19ikr.  lOA.  3R.  36P.,  take  7ikf.  325^.  13P. 

2.  From  6^.,  take  2R.  16P. 

3.  From  Idsq.yd.  5sq.in.,  take  ISsq.yd.  143sq.i7t, 

CUBIC    MEASURE. 

1.  From  320C.yd.  20C.ft.,  take  3090.  y^.  21C./^  l,400a»». 

2.  From  l5cords  lOOC.ft.,  take  lO^^or^?^  116C./if. 

3.  From  2iC.yd.,  take  65C./^.  895C.^V^. 

WINE    MEASURE. 

1.  From  eT.  Ihhd.  39gaL  2qt.,  take  2T.  45gaL  Iqt, 

2.  From  IQgal.  Iqt.  Ipt.  2gL,  take  \5gal.  3gL 

3.  From  i5gal.j  take  15qt.  2pt.  \gi. 

DRY   MEASURE. 

1.  From  34c7i.  265?^.  7?^,  take  16c7i.  26^2^.  3;?/%  3j^  \pt. 

2.  From  38i?*.  7^^^/.  3qt.  1;;^.,  take  3Uw.  Ij)^. 

3.  From  IQbu.^  take  2^^.  \pt. 

TIME. 

1.  From  30yr.  29da.  40",  take  28^7*.  6?7Jo.  Shr,  51'  45". 

2.  From  lOyr.  Gmo.,  take  15<r/«.  6w. 

3.  From  lyr.,  take  29Ar.  17'  35". 


DENOxMINATE   FRACTIONS.  223 

CIRCULAR   MEASURE. 

1.  From  50°  26'  50",  take  49^'  31'  55". 

2.  From  9S.  25^,  take  51*^  51". 

3.  From  315°,  take  59'  35". 

N.  B. — In  the  sequel,  examples  illustrative  of  each  lesson  will 
be  given  at  the  end  of  it.  But  questions  relative  to  it,  or  com- 
bining it  with  others,  will  be  found  in  the  Appendix,  for  more 
advanced  students,  revising  arithmetic.  Those  included  in  the 
lessons,  are  for  beginners. 

LESSON  LIV. 

ADDITION  AND  SUBTRACTION  OF  DENOMINATE 
FRACTIONS. 

1.  Fractions  of  different  denominations  of  the  same 
measure  have  frequently  to  be  added  or  subtracted  3  as, 
for  example, 

•|  of  a  pound  and  ^  of  a  shilling. 
All  that  is  necessary  is  to  reduce  them  to  the  same 
denomination  j  which  may  be  done  in  several  ways,  either 

2.  By  transforming  the  fractions  into  integers  of 
smaller  denominations. 

By   the   rule   of  Lesson   L.,   we   may,   for  example, 

change  ....         £^     into     6sh,    8d. 

and  ....         ^sh,  into     Osh.    Sd. 


Their  sum,  under  this  form,  would  be     .         6sh,  lid 
and  their  difference  .  .  .  6sh.    5d. 

,     3.  Or  else,  by  transforming  the  fractions  into  others 
having  all  the  same  denomination. 

By  the  rule  of  Lesson  XLIX.,  we  may  change  £-j- 
into ^-^sh,  =  ff . 


Which  may  be  added,  as  usual,  to  \  sh. 


3 


The  sum  being      ....  41^^* 

Or,  we  may  change  the  second  fraction  into  the  frac- 
tion of  a  pound,  by  making  it       .         .         £-^q  =  gf^. 
Which,  added  to      .         .         .         .         £1=2%. 


would  give  ....  £A^?j' 


224  LESSON    LIV. 

These  i\.^ee  sums,  6sh,  lid.;  f  |s/t. ;  £^Vu?  ^^^  ^^^^' 
valent. 

The  form  to  be  preferred  depends  on  the  nature  of  the 
question  proposed.  The  corresponding  differences  in 
fractional  form  would  be         ...         .         5|5A. 

Op  :^  7  7 

Both  equivalent  to  6sh,  5d. 

4.  If  the  fractions  were  decimal,  instead  of  vulgar,  the 
operation  would  be  similar.     Thus, 

OMlb.  and  ,75dwt. 

might  be  added  as  follows : 

OPERATION. 

0.65x20=  iSdwt. 
0.75 


ISJMwt. 
Or,  because  .15dwL  =  18gr,y  it  would  also  be  ISdwt,  + 
ISgr. 
The  subtraction  between  the  same  would  give 

12.25dwt. 
Or,  .         .         12dwf„  6gr. 

These  are  mere  applications  of  what  has  been  taught 
in  Lessons  XLIX.  and  L. 

Questions. — How  do  you  add  fractions  of  the  same  measure, 
but  of  different  denominations  ?  How  do  you  subtract  them  ? 
What  determines  the  form  to  be  chosen  ?  How  do  you  change 
fractions  into  integers  of  lower  denominations?  How  do  you 
transform  a  fraction  into  another  of  a  higher  denomination  ?  Of 
a  lower  denomination  ? 

EXERCISES   IN    THE    THREE    FORMS. 

1.  £i  +  iV^-  =  ^5sh.  lOyV-  =  ^U-^^'-y    or  £{iU* 

2.  %yd,^ift.-^lnil.=z  l,5i0yd,2ft.9{7i.,  or,  ^  .  or    decimally, 

^.  lwl:.-^\da,-\-\hr,  =  2da,\A\hr,  or,  |  .  or         « 

o 
4.  £^  —  Yosh.=  I5sh,  S^^d.  or,|  -or         '^ 

6,  Xf  —  £5^.=  145^.  3d.  or,"^  .  or         «« 


COMPOUND    MULTIPLICATION.  22"5 

6.  -l-yd,  —  ^in.  =  5|m. 

7.  £%J^^sk.=  I3sh.  lOd,  2% far.,  or  J-f^^A.,  or  decimally  ?  .  . 

8.  Iwl^  —  9j Qda.  =  5w^,  Ada.  Ihr.  12m. 

9.  £^  4-  ^sh.  +  j\d.  =  3sh.  Id.  iff /«r.,  or  m^sJi.,  or 
decimally?  .... 

10.  £|-  —  ■§  of  ish.  =  lOsh.  Id.  \\far.,  or  VV^^-s  or ? 

11.  f  of  £5j-— |.sA.  =  i;i  S^A.  IIJ.  OJI/ar.,  or  £\\%'^^ 
or ? 

12.  What  will  remain,  if  \3\sh.  and  Tfc/.  be  taken  from  jCI  ? 

13.  Which  is  greater,  J  of  ^sh.  or  l^A.  2|i^.  ? 

^;i5.  1^^.  2|fi?.  by  yV^« 

14.  irtt^if.  +  15/i.  +  \cxut.  +  9o^.  +  \lh.  =  3^r.  25/5.  10|o^. 

15.  \oz.  —  %dwt.  —  Idwt.  9^r.,  or  (fraction)  or  .  .  (decimal). 

16.  £0.5+  1.255/i.  4-0.12^.=  115;^.  3.12J.,  or  135.12^. 

17.  0.16c^d;^  +  2^r.  +  .75/5.=  2^r.  18.67/i. 

18.  0.75w/.  +  0.9P.  +  3.25y6i.  =  241P.  2yd.  2.\ft. 

19.  X0.75  — .905^.=  145^.  Id.  0.8/ar. 

20.  0.35/5.  —  .85o^.  (troy)  =  3oz.  Idwt, 

21.  0.65^.  •—  21.75P.  =  82.25P. 

LESSON  LV. 
MULTIPLICATION  OF  COMPOUND  NUMBERS. 

1.  Compound  numbers  being,  in  reality,  only  mixed 
numbers,  we  know  already  a  method  by  which  they 
might  be  multiplied.  For  this  purpose,  it  would  suffice 
to  put  the  lower  denominations  under  the  form  of  frac- 
tions )  and,  indeed,  in  many  cases,  this  might  be  the  best 
way. 

2.  If  we  had,  for  instance,  to  multiply 

£7  Ibsh.    by    \ft.  Sin., 
we  might  change  them  into     £7|     and  ^ifi* 

Or,  decimally,  £7.75       and    1.25/j?., 

and  then  multiply. 

But,  in  general,  this  is  not  the  most  expeditious  method. 

3.  The  operation,  as  usually  performed,  is  much  more 

P 


226  LESSON   LV. 

difficult,  though  shorter.  It  requires  much  attention,  and 
can  be  explained  clearly  only  by  examples.  But,  before 
giving  these,  it  is  important  to  present  some  remarks  on 
the  nature  of  multiplication. 

4.  Multiplication  is  the  addition  of  a  quantity  to 
itself  a  certain  number  of  times ;  so  that,  whatever  the 
multiplicand  may  be,  the  multiplier  is  altvays  made  an 
abstract  number,  and  the  product  is  invariably  of  the 
same  nature  as  the  multiplicand  (XL,  7). 

5.  It  is  therefore  a  great  absurdity  to  speak  of  multi- 
plying together  denominate  quantities ;  as  pounds  or  doU 
lars  and  yards.  When,  for  instance,  the  cost  of  so 
many  yards  of  cloth,  at  so  much  a  yard,  is  required,  v^^e 
do  not  multiply  money  by  yards;  but  we  take  the  value 
of  a  yard  as  many  times  as  there  are  units  in  the  number 
of  yards,  and  the  amount  is  a  sum  of  money. 

Thus,  4  yards  of  cloth,  at  3  dollars  a  yard,  are  3  dollars  taken 

4  times ;  that  is,  12  dollars. 

Again:  if  one  dollar  will  buy  a  certain  number  of 
yards,  and  we  wish  to  know  how  many  yards  we  can 
buy  with  a  certain  sum  of  dollars,  we  repeat  the  number 
of  yards  as  often  as  there  are  dollars  in  this  sum ;  and  the 
amount  is  a  number  of  yards,  with  which  the  name  of 
the  unit  we  multiply  by  has  nothing  to  do. 

Let  it  be  supposed,  for  instance,  that  4  yards  cost  1  dollar;  if 
it  be  asked  how  many  yards  can  be  had  for  3  dollars,  we  merely 
repeat  4  yards  3  times ;  the  denomination  dollar  having  nothing 
to  do  with  the  result. 

6.  There  is  one  seeming  exception  to  this,  namely : 
when  we  multiply  yards  by  yards,  feet  by  feet,  or  any 
other  measure  of  length  by  itself,  and  get  square  yards, 
square  feet,  or  some  other  square  measure.  But  this,  in 
reality,  is  a  mere  short  mode  of  expression,  which  should 
not  be  understood  literally. 

Suppose,  for  instance,  that  we  have  a  square  board, 

5  inches  in  length  by  6  in  breadth,  and  we  wish  to  ascer- 
tain how  many  square  inches  its  surface  contains ;  we 
divide  its  sides  into  inches,  and,  by  drawing  lines  across, 


COMPOUND    MULTIPLICATION. 


227 


1 

2 

3 

4 

5 

30 

form  squares,  each  one  square 
inch ;  and  then,  instead  of 
counting  each  square,  we 
count  the  inches  on  both 
edges,  and,  multiplying  them 
together,  call  their  product 
30  square  inches. 

This,  however,  if  pro- 
perly investigated,  is  not 
multiplying  linear  inches  by 
linear  inches.  But  we  ob- 
serve that,  along  the  upper 
row,  there  are  as  many 
square  inches  as  there  are 
divisions  in  it  j  and,  also, 
that  there  are  as  many  such 

rows  as  there  are  inches  along  the  other  side.  Conse- 
quently, we  repeat  the  number  in  one  row,  as  many  times 
as  there  are  rows,  which  is  here  5  multiplied  by  6. 
This  evidently  is  not  multiplying  inches  by  inches,  but 
repeating  a  number  of  small  squares,  equal  to  that  of 
the  inches  along  one  line,  by  a  number  of  units  equal  to 
the  number  of  inches  on  the  other  line. 

7.  Nor  is  it  correct  to  consider  the  multiplication  of  feet 
by  feet,  as  producing  always  square  feet ;  it  may  some- 
times give  linear  feet,  as  some  subsequent  examples  will 
show  (LVIII.).  In  this,  as  in  other  cases,  a  precise  state- 
ment alone  can  determine  the  nature  of  the  product. 

8.  We  cannot  properly,  therefore,  multiply  one  com-  | 
pound  number  by  another.  If  we  were  asked,  for  exam- 
ple, to  multiply  Sib.  9oz.,  troy,  by  7^^.  6d.,  the  question 
would  be  undetermined  ;  for  nothing  tells  us  which  is  the 
multiplicand  or  the  multiplier.  And  even  if  either  num- 
ber was  designated  as  the  multiplicand,  we  do  not  know  to 
what  unit  of  the  multiplier  it  is  equal.  To  illustrate  this 
fully,  we  will  frame  four  different  questions  upon  the  same 
example  : 

1st.  If  one  pound  cost  7sh,  6d.,  what  will  be  the  cost 


228  LESSON   LV. 

of  sib,  9ozJ  Here  the  muHiplIcand  is  75^.  6d.,  and  the 
unit  it  is  equal  to  is  1  pound  weight. 

2d.  If  one  ounce  cost  7sh.  6d.,  what  will  be  the  cost 
of  Sib.  9oz.  ?  In  this  case,  the  multiplicand  is  the  same, 
but  it  is  equal  to  another  unit,  1  ounce,  and  the  product 
will  be  12  times  larger. 

3d.  If  Sib,  9oz,  cost  1  shilling,  how  much  can  be 
bought  for  7sh,  Gd.'^  In  this  third  case,  the  multiplicand 
is  8lb,  9oz.y  and  the  unit  of  the  multiplier  is  1  shilling. 

4th.  If  Sib,  9oz.  cost  1  penny,  how  much  can  be 
bought  for  7sh,  6d,  ?  In  this  fourth  question,  the  multi- 
plicand is  the  same  as  in  the  preceding,  but  it  is  equal  to 
the  smaller  unit,  1  penny,  and  the  product  is  larger. 

It  is  obvious  that  each  of  these  questions  will  furnish 
a  different  answer.  The  first  two  will  give  pounds,  shil- 
lings, and  pence,  in  the  product;  the  last  two,  pounds 
and  ounces. 

9.  It  must  be  clear,  therefore,  that  we  cannot  proceed, 
in  any  case,  to  multiply  denominate  numbers,  unless  it  be 
distinctly  understood  which  unit  of  the  multiplier  the 
multiplicand  is  equal  to.  Then,  the  number  of  those 
units,  considered  as  an  abstract  number,  constitutes  the 
multiplier. 

This  remark  does  not  apply  to  pounds,  shillings,  and 
pence,  alone ;  it  is  equally  applicable  to  dollars  and  sim- 
pler numbers.     If  it  were  proposed,  for  instance,  to  mul- 
tiply 4i  yards  by  $1.25,  the  question  would  be  fully  as 
,  incomplete ;  for,  one  dollar^  one  dime,  or  one  cent,  may 
i  be  the  value  of  the  yard ;  or  it  may  be  either  the  yard 
]  or  the  quarter  which  is  worth  $1.25.     Therefore,  to  any 
such  question,  the  pupil  must  answer  by  asking  for  a  clear 
statement,  by  which  he  can  ascertain  his  data. 

10.  I  have  enlarged  upon  this  point,  because  import- 
ant mistakes  are  frequently  made  in  compound  multipli- 
cations, for  want  of  due  consideration,  as  regards  the 
unit  of  comparison;  and  I  have  often  heard  such  a  ques- 
tion as  this  propounded:    What  is  the  product  of  £2 


COMPOUND    MULTIPLICATION.  229 

195^.  6 J.  hy  £2  l^sh,  ^dA  and  seen  wonder  expressed 
that  the  answers  given  by  different  persons  did  not  agree. 

Persons  conversant  with  the  correct  principles  of 
arithmetic,  will  not  discuss  seriously  such  a  question, 
which  is  indefinite,  even  with  dollars. 

Let  it  be  proposed,  for  instance,  to  multiply  $1,11  by 
$1.11:  the  answer  may  be  $123.21;  $12,321;  or 
$1.2321,  or  even  other  numbers,  according  to  the  ques- 
tion, which  might  be, 

1st.  A  man  has  adventured  $1.11,  and  made  $1.11  for  each 
cent :  how  much  has  he  made  in  all  ?  Aiis.  $123.21. 

2d.  His  gain  has  been  $1.11  for  each  dime.  Ans.  $12,321. 

3d.  His  gain  has  been  $1.11  for  each  dollar.         A7is,  $1.2321. 

This  is  sufficient  to  show  what  a  great  variety  of  an- 
swers may  be  obtained  by  changing  the  unit  of  compari- 
son. The  intelligent  pupil,  therefore,  should  not  accept 
as  a  question  a  multiplication  so  incompletely  stated. 

11.  In  fact,  multiplication  is  always  the  consequence 
of  a  species  of  rule  of  three  ;  in  which  one  of  the  terms 
is  one  unit,  of  the  nature  of  the  multiplier ;  so  that,  in 
general,  the  statement  would  be, 

1  :  the  multiplier  :  :  the  multiplicand  :  the  product. 

For  example : 

07ie  yard  has  cost  4  dollars :  how  much  will  3  yards  cost  ? 
would  give  the  proportion 

Zyd, 
\yd.  :  2yd,  :  :  $4  :  the  answer  =  $4  X  yIT 

That  is,  the  answer  would  be  $4,  multiplied  by  the 
ratio  of  3  yards  to  1  yard ;  which  is  nothing  but  the  ab- 
stract number  3,  the  nature  or  denomination  of  the  num- 
bers havino;  nothino;  to  do  with  their  ratio.  This  shows 
why  the  multiplier  is  always  an  abstract  number. 

If,  in  all  cases,  this  is  kept  in  view,  multiplications 
indefinitely  stated  as  the  above,  will  no  longer  be  re- 
garded as  arithmetical  puzzles. 

The  question  is  incomplete  until  the  unit  to  which  the 
multiplicand  is  equivalent  is  given. 
20 


230 


LESSON    LV. 


12.  The  general  'principle  of  compound  multiplication 
is  to  multiply  successively  each  part  of  the  multiplicand 
by  each  part  of  the  multiplier;  having  regard,  of  course, 
to  the  dependence  of  the  various  orders  of  units. 

We  will,  however,  consider  three  cases. 


13.  When  the  multiplier  is  a  small  simple  number: 

Let  it  be  proposed,  for  example,  to 

multiply .£247  17 sh.  lid. 

by         ....         .  9 

The  product  will  be  .     <£2,231    Ish.    3d. 

In  this  case,  you  multiply  severally  the  parts  of  the 
multiplicand^  beginning  with  the  lowest  denomination^ 
and  carry  successively^  to  their  proper  column^  the  inte- 
gers  of  higher  denominations^  furnished  by  the  inferior 
products. 

Thus,  the  product  of  11  <?.  by  9,  is  99cZ.,  equal  to     Ssh,  3J. ; 

you  set  down  3(Z.,  and  carry  85/z.  to  their 

proper  column. 
Then,  9  times  11  sh,  are  1535^.,  and  8  carried, 

161s/i.  =£8  1s^.; 

you  set  down  1  s7i.  and  reserve  j£8  to  be  carried. 
Finally,  9  times  £7,  and  £8  carried,  are  £71,  &c. 

Questio7is. — How  is  the  multiplier  considered  in  all  multiplica- 
tions ?  Can  you  multiply  dollars  by  yards  ;  yards  by  dollars ;  or 
any  denomination  by  another  ?  What  is  meant  when  feet,  multi- 
plied by  feet,  are  said  to  make  square  feet  ?  Is  the  question 
definite,  when  it  is  asked  to  multiply  a  compound  number  by 
another  ?  What  is  required  to  make  it  definite  ?  How  do  you 
multiply  a  compound  number  by  a  simple  small  number?  Is 
there  any  analogy  between  multiplication  and  the  rule  of  three  ? 

EXERCISES. 

1.  How  much  are  8/5.  of  tea,  at  5sh.  S^d.  ?     Ans.  £2  5sh,  Sd. 

2.  «  9r7vt.  of  cheese,  at  £1  llsk.  5d.  per  ewt,^ 

Ans,  £U  2sh,  9d.  . 


COMPOUND    MULTIPLICATION. 


231 


3.  How  much  are  12cv;t.  of  sugar,  at  ^3  Ish.  4J.  per  cwt.  1 

Alls,  X'40  Ssh 

4.  Multiply  14/5.  ^oz.  Udwt,  llgr.  by  5. 

Ajis.  74/5.  Oo^.  I3dwt,  IS^r. 


«         2^a.  6/ir.  SP^z.  by  19. 


.4?i5.  A^da.  ^hr,  30w. 


Thomas  Leslie 


Richmond,  19th  May,  1846. 
Bought  of  John  Richardsy 


If  pounds  of  tea,     .  at     Ash.  6d,     . 

4  bushels  of  corn,  at ,   5sh.  4d, 

5  quarts  of  brandy,  at     8=5^.  4d.  per  gallon. 

6  do.      rum,      .  at     Ish.  6d.  per  gallon. 

7  J  yards  of  chintz^  at     5sh.  5d. 


£0    Ish.  IQU, 


£6  \^sh.    4f^. 


Charleston,  2d  January,  1847. 


William  Hunt 

Bought  of  James  Nichols, 

2    pieces  muslin, 

at 

2Qsh. 

£3    Osh.  Od. 

25    yards  linen. 

at 

2sh, 

28|  yards  calico, 

at 

2sh.  Q>d. 

28|  yards  flannel. 

at 

2sh,  2d, 

1    piece  bombazette, 

at 

5Qsh, 

2    pieces  blue  calico, 

at 

51sh,  ed. 

50|  yards  dimity. 

at 

2sh.  ed. 

3    pieces  drilling, 

at 

S4sh. 

£39  ll^A.  2d. 

LESSON  LVI. 

CASE  II. 

1.  When  one  of  the  factors,  though  simple,  is  a  large 
number,  the  reduction  of  the  inferior  products  cannot  be 
effected  mentally.  In  that  case,  the  method  of  the  first 
case  is  not  the  most  expeditious.  The  operation  will  be 
more  readily  performed  by  means  of  aliquot  parts,  com- 
monly called  Practice. 


232  LESSON   LVI. 

MULTIPLICATION  BY  ALIQUOT  PARTS,  OR 
PRACTICE. 


2.  Let  it  be  proposed  to  multiply 
by 


for  lOsh, 
for     5sh, 
for     bd, 
for     3d. 


OPERATION. 

£784 

15sk.    9d. 

857 

£5488 

3920 

N 

6272 

428 

lOsh, 

214 

5sh, 

21 

8sh,    6d. 

10 

Usk,    3d. 

£672,562     llsh.    9d. 

Here,  after  having  multiplied  the  whole  numbers  toge- 
ther, as  usual,  we  pass  on  to  the  multiplication  of  the 
fractional  part,  15  shillings  by  857. 

In  order  to  this,  we  separate  the  number  of  shillings 
into  the  most  convenient' aZi(/i/o^  parts  of  a  pound,  which 
here  are  10  and  5,  and  multiply  separately  10  shillings 
and  5  shillings ;  the  sum  of  the  two  products  being  evi- 
dently the  same  as  that  of  15  shillings. 

Now,  the  product  of  1  pound  by  the  multiplier  being 
£857,  it  is  evident  that  the  product  of  half  a  pound 
must  be  one  half  of  £857.     So  that  we  have, 

For  lOsh.         .         .        £4.28  lOsh. 

And  because  5  shillings  are  one  half  of  10  shillings,  the 
product  of  5  shillings  by  857,  must  be  one  half  of  that 
of  10  shillings.     Hence  we  get  at  once, 

For  5sh.  .         .         £214  5sh. 

Having  now  completed  the  multiplication  of  the  shil- 
lings, we  pass  on  to  that  of  the  9  pence.  These  we  also 
decompose  into  the  most  convenient  aliquot  parts,  namely, 
6  and  3. 

6  pence  is  one  half  of  a  shilling ;  consequently,  J^j  of 
5  shillings.  Hence,  the  product  of  6  pence  by  857, 
must  be  -^^  part  of  the  product  of  5  shillings.  Thus 
we  get 


COMPOUND    MULTIPLICATION. 


23a 


For  6d. 


£21  SsL  6d. 


Finally,  the  product  of  3  pence  is  one  half  of  the  pro- 
duct of  6  peaQG  by  857  •  which  is. 


For  3d, 


<£10  Usk.  9d. 


This  method  is  expeditious  and  simple.  Each  indivi- 
dual product  is  made  to  depend,  by  a  simple  relation  of 
aliquot  numbers,  upon  some  preceding  product.  As  re- 
gards the  choice  of  the  most  convenient  aliquot  parts, 
and  the  progress  of  dependence  of  the  successive  pro- 
ducts^ no  positive  rule  can  be  given  j  practice  alone  will 
make  it  familiar. 

3.  As  a  second  example,  let  us  propose  this 

Question. — At  65Z5.  14^oz.  13di\  for  «£1,  Jiow  much 
can  be  had  for  £59  ? 


operation. 

6516,     Uoz, 
59 


13dr, 


I  of  59 

J  of  last 
1 

2 

A  of  last 

1 

1 
4 


585/6. 

325 

29       So2. 

. 

for  Soz 

14     12 

. 

for  4. 

7       6 

, 

for  2. 

1     13 

8dr, 

for  Sdr, 

0     14 

12     . 

for  4. 

0       3 

11     . 

for  1. 

3,889/3.  90-2:.  I5dr, 


After  having  multiplied  the  two  integers  together,  we 
decompose  14oz.  into  8,  4,  and  2,  and  multiply  them 
severally  by  the  multiplier,  as  shown  in  the  operation. 

Then  we  pass  on  to  the  ISdr,,  which  we  decomposB 
into  8,  4,  and  1.  And  then,  because  Sdr.  is  ^oz.,  it  is 
i  of  2oz. :  we  therefore  take  l  of  the  product  of  2oz,, 
and  proceed  with  the  other  aliquot  parts  according  to 
their  relative  values. 

4.  If  the  question  had  been  : 
20* 


234 


LESSON   LVI. 


OPERATION. 
£59 

e5lb.   Uoz.  13dr, 


The  cost  of  lib,  being  £59,  what  icill  be  that  of 
65lb.  Uoz,  ISdr.?  -^ 

The  multiplicand  would  no  longer  be  the  compound 
number  5  it  would  be  £59,  and  the  product  must  be  in 
£  sh,  and  c?.,  but  the  operation  would  be  nearly  the 
same. 

After  having 
multiplied  the  two 
integers  together 
as  usual,  we  de- 
compose the  frac- 
tional parts  in  the 
same  way  as  be- 
fore, and  get  for 
Soz,  one  half  of 
£59,  that  is  £29 
lOsh.  Then,  the 
other  subdivisions 
depending  on  this, 
according:    to   the 


£585 

325 

for  Soz, 

29 

lOsk, 

4 

14 

15 

2 

7 

7       6d, 

16 

8clr, 

1 

16     10| 

.       8 

4 

•      0 

18       5i 

.      4 

1 

0 

4      7/, 

.      5 

X3,889     12sk.  5j\d. 


J  ^  —  1  1 
16"—  ^i'e 

same  scale  as  in  the  preceding  question,  give  the  results 
set  down  in  the  operation. 

5.  In  these  operations  it  will  be  observed  that  the 
nature  of  the  multiplier  is  not  considered,  but  merely 
the  abstract  numerical  relation  of  its  parts. 

6.  Sometimes  it  may  even  be  preferred  to  multiply  in 
this  way  by  a  fraction. 


Let  it  be  proposed  to  multiply 

by 

We  have  for  y^ 
for    4   . 
for    1   •  •  • 


OPERATION. 


H 

29.50 
14.75 
3.56i 

7.81A 


Questions. — ^When  one  of  the  factors  is  a  large  simple  number, 
how  do  you  multiply?  What  are  aliquot  parts?  How  are  the 
integers  multiplied  ?  How  the  smaller  parts  ?  What  aliquot 
parts  should  be  preferred  ?  How  do  you  ^orm  individual  pro- 
ducts from  others  1     If  the  single  number  is  the  multiplicand, 


COMPOUND    MULTIPLICATION,  235 

does  the  operation  differ  materially  ?  Are  thft  denominations 
considered  in  the  multiplier?  How  are  they  considered ?  Can 
you  multiply  with  fractions  by  aliquot  parts  ? 

EXERCISES. 

1.  How  much  are  53  loads  of  hay,  at  £3  15.s/i.  2d.  per  load  ? 

Aus.  £199  3sh.  lOd, 

2.  «  79  bushels  of  wheat,  at  lUh.  5f<i.  per  bushel  ? 

A?is.  £45  6sk.  lO^d, 

3.  "  94  casks  of  beer,  at  12sh.  2d.  per  cask? 

A71S.  £51  3sh.  Sd. 

4.  "         109  barrels  of  fish,  at  Us/i.  6d.  per  barrel  ? 

Ans.  £19  QsJu  Qd, 

5.  "         345  barrels  of  pork,  at  £1  3sh,  9d.  ? 

Ans,  £409  13sh,  9d. 

6.  "  47  casks  of  rice,  each  weighing  2cwt.  \qr.  23lb.  ? 

Ans,   \\5cwt,  \qr,  lllb. 

7.  In  the  lunar  cycle  of  19  years,  of  3Q5da,  5hr.  48'  49",  how 
many  days  and  parts  of  a  day  ?  Ans,  6,939^«.  \Ahr,  27'  31 '. 

8.  How  much  water  will  be  contained  in  57  cisterns,  each 
containing  455gal,  3qt,  lpt,l  Ans.  25,984^«/.  3^^.  Ipt, 

9.  A  father  gives  each  of  his  7  sons  150^.  3R.  12F. :  how 
many  acres  in  all  ?  Ans.  1,055^.  3R,  4P. 

10.  What  is  the  whole  weight  of  11  ingots  of  silver,  each 
weighing  416.  loz,  l5diot,  22gr.l        Ans.  A5lb,  loz,  15dwt,  2gr. 

11.  At  $1.22|  an  ounce,  what  will  be  the  cost  of  a  vase  weigh- 
ing 3lb,  5oz,  lldwt,  19^r.?  Ans.  $51.3147_l., 

12.  At  the  rate  of  I2dwt.  Igr,  for  $1,  of  what  weight  could  a 
vase  be  made  for  $148?  Ans.  lib.  6oz.  19dwt,  Agr, 


LESSON  LVII. 

CASE    III. 

1.  The  last  case  is  when  both  the  multiplicand  and 
multiplier  are  compound  numbers. 

Let  us  begin  with  the  simple  example  of  a  fractional 
multiplier. 

1st  Question. — A  yard  of  a  certain  work  has  cost 
j£65  17 sli.  lid.:   how  much  will  39|^  yards  cost? 


236 


LESSON    LVII. 


OPERATION. 

£65     11  sh,  lid. 

39| 


For  lO^A. 

5 

2 
For    6^. 

3 

2 

For    pd, 

2 

5 


£585 

195 

19 

IQsh. 

9 

15 

3 

18 

0 

19 

6d, 

0 

9 

9 

0 

6 

6 

32 

18 

111 

16 

9 

5| 

8 

4 

81 

£2,627     llsh,  l\\d. 


17    =  Oi 

g      —   ^8 


We  multiply,  first,  by  the  integer,  39,  as  in  the  pre- 
ceding case,  which  gives  the  first  eight  partial  products. 

There  remains,  then,  to  multiply  by  | :  we  decompose 
•J-  into  I,  |,  and  -i,  and  observe  that,  since  the  product 
by  1,  would  be  jG65  11  sh.  11^.,  the  product  by  |  must 
be  one  half  of  it  j  so  that  we  get, 


For 

Then,  for 

And,  finally, 

For 


one  half  of  this,  or 


£32  18sA.  IV^d. 
£16     95/t.    bid. 


<£8       45/i. 


^d. 


one  half  again,  or 

which  is  the  last  partial  product.     The  sum  of  all  the 
partial  products  is  the  answer. 

In  this  example,  we  might  have  multiplied  by  40,  and 
subtracted  the  product  by  |. 

2.  2d  Question. —  The  yard  of  a  certain  piece  of 
work  has  cost  £25  19sh,  5d,:  what  will  69yd.  2ft,  llin* 
cost  ? 


COMrOUND    MULTIPLICATION. 


237 


OPERATION. 

£25      19  sh,    5d, 
myd.  2ft,  Win. 


£225 

150 

For  \Qish. 

• 

34 

lOs/i. 

5 

,         , 

17 

5 

4  (J 

of  69) 

13 

16 

For  5^.(^32 

of  product  by  5jA.)     1 

8 

9d. 

36 

For  Ift, 

,         , 

8 

13 

]f 

.    24 

1      . 

•         • 

8 

13 

If 

.    24 

For  6t7i, 

•         • 

4 

6 

n- 

.    30 

3      . 

•         • 

2 

3 

^h 

.    15 

2      . 

•         • 

1 

8 

10t\ 

.    10 

£1,817 


Ash. 


^3  6" 


103 
3B" 


—  ^^-^ 


In  this  example,  the  yard  is  the  unit,  equal  to  the  mul- 
tiplicand. 

We  multiply,  first,  the  cost  of  1  yard  by  69,  as  in 
Case  II.  We  will  only  remark  that,  5  pence  being  the 
twelfth  part  of  5  shillings,  we  take  y^^  of  this  product. 

Then  we  come  to  the  fractional  parts  of  a  yard ;  and, 
since  1  foot  is  ^  of  a  yard,  we  take  J  of  the  value,  j£25 
19sA.  5c£.,  of  a  yard,  and  get. 

For  Ift.         .        .         £S  13sh.  Ift?., 
which  we  repeat  a  second  time. 

Now,  for  6in,  we  take  I  of  this,  or       £4^    6sk. 
Then,  for      3,         .  J  again,    or  2    3 

And,  finally,  for  2,  }  of  the  product  by  6,  or  1 
And  then  add  up  the  partial  products. 

3.  If  the  multiplicand  was  the  value  of  1  foot  instead 
of  a  yard,  the  product  would  be  three  times  as  much. 
The  69  yards  must  be  first  changed  into  feet. 

Ans,  je5,451   Ush.  2j\d. 

4.  If  it  was  the  value  of  1  inch,  both  yards  and  feet 
should  first  be  converted  into  inches ;  and  the  product 
would  be  36  times  larger.  Ans,  .£65,420  lOsh,  Id. 


S88  LESSON   LVII. 

5.  It  might  also  be  the  cost  of  a  rod.  In  this  case,  the 
product  would  be  still  different ;  the  number  of  rods  in 
69  yards  must  be  extracted,  and  then  the  product  would 
be  found,  in  a  similar  way,  to  be  .£330  8sh.  l-\}, 

6.  These  different  suppositions,  which  I  propose  as 
exercises,  show  that,  in  compound  numbers,  we  cannot 
propose  simply  to  multiply  two  compound  numbers,  but 
must  designate  the  unit  of  value. 

7.  As  a  last  example,  let  us  take  a  question,  in  which 
the  unit  of  value  is  larger  than  the  whole  multiplier. 

3d  Question. — Sctvt.  Sqr,  21lb,  cost  £1  :  how  much 
can  be  had  for  llsh,  Id.l 

operation. 

For  £1  ;         .         ^cwt,    Sqr.  21lb. 

:eO         llsh.    Id, 


For  lOsk.    . 

Icwt, 

3gr.21lb,    8oz. 

1 

0 

0       22        5          9jdr. 

For  6d. 

0 

0       11        2        12| 

1 

0 

0         1      13        12| 

2  1         6      14  31 

5 

Here,  the  value  of  £1  being  the  multiplicand,  the 
value  of  10  shillings  is  one  half  of  it;  and,  from  this 
first  product,  we  obtain  ail  the  others,  as  the  operation 
shows. 

8.  Operations  so  complicated  are  seldom  met  with ; 
but  they  are  good  exercises,  depending  almost  exclusively 
on  the  judgment  of  the  operator,  and  in  which  he  cannot 
proceed  by  blind  rote.  As  such,  they  are  well  calculated 
to  improve  the  readiness  of  the  student  in  figures,  and 
his  quickness  in  the  combination  of  different  units. 

PROOF. 

9.  The  simplest  way  to  prove  compound  multiplica- 
tion is  to  double  either  multiplicand  or  multiplier,  and 
proceed  thus  wdth  the  operation,  which  must  give  a 
double  product. 


i>t;odecimals.  239* 

The  proof  by  division  would  generally  be  too  com- 
plicated. 

Qucstio7is. — How  do  you  multiply  two  compound  numbers 
together  ?  What  parts  do  you  multiply  first  ?  Is  it  sufficient  to 
give  the  multiplicand  and  multiplier  ?  What  else  must  be 
known?  Give  examples.  How  do  you  proceed,  when  the  unit 
of  value  is  an  inferior  order  of  the  multiplier  ?  When  it  is 
superior  to  the  orders  in  the  multiplier  ?  Give  examples.  What 
is  the  simplest  mode  of  proving  compound  multiplication  ?  Could 
it  be  proved  by  division  ? 

EXERCISES. 

1.  If  one  rod  of  a  certain  work  costs  25£  l%sh,  5d.,  how 
much  will  Q9yd.  2ft.  Win,  cost?  Ans.  £330  85A.  \\\d. 

2.  If  one  foot  costs  £25  X'^sh.  5d.,  how  much  will  69yd,  2fe, 
11m.  cost?  A71S.  £5ji5l  Ush.  2^.2^. 

3.  If  one  inch  costs  £25  19sh.  5d.^  how  much  will  69yd.  2ft, 
lltVj.  cost?  ^«5.  £65,420  105/i.  7(f. 

4.  If  one  yard  costs  £3  2sh.  6d.^  what  will  5yd.  2ft.  Sin. 
cost?  A?i.s.  £17  I9sh.  4|rf. 

5.  A  star  describes  an  arc  of  2°  35'  in  a  year :  what  arc  will 
it  describe  in  21yr.  9mo.  ?  Aiis,  71°  41'  15". 

6.  In  a  certain  business,  the  capital  has  produced  £15  11-5^. 
5d:  for  each  pound  invested :  how  much  must  a  partner  receive 
whose  share  is  £31  175/^.  9d.  ?  A?is.  £496  lOsh.  3|7^. 

7.  At  6sk.  Bd.  an  ounce,  what  will  a  vase,  weighing  2lb.  lOoz, 
Udwt.  cost?  A71S.  £11  llsk.  4d. 

8.  What  is  the  tax  on  £745  Usk.  Sd.,  at  3sk.  Qd.  in  the 
'pound?  A?2s.  £130  lOsh.  Od.  3± far, 

9.  How  many  hundred  weight  of  raisins  in  7A  casks,  each 
containing  2cwt.  9qr.  25/5.?  Ans.  IQcwt.  3qr.  \6lb.  Q^oz, 

10.  How  many  pounds  of  coffee  in  13|  bags,  each  containing 
Icwt.  3qr.  15/5.  ?  Ans.  25cwt.  3qr.  17/5.  4oz. 


LESSON  LVIII. 

MULTIPLICATION  OF  MEASURES  OF  LENGTH,  AND 
DUODECIMALS. 

1.  For  some  practical  purposes,  as,  for  instance,  to  mea- 
sure work,  the  foot  is  divided  into  12  inches  or  primeSy 


240 


LESSON   LVKI. 


OPERATION. 

24/if. 

r   6" 

by 

2ft. 

48 

5'     4" 

For  Qiiu 

.  1 

For  1      . 

0 

2m. 

For  Qsec. 

.  0 

1 

For  4i>i. 

8 

2      6.<?ec. 

For  1      . 

2 

0      7        6"^ 

For  4"    . 

0 

8      2        6 

marked  ';  the  inch  or  prime  is  subdivided  into  12  seconds, 
marked  '';  the  seconds  into  12  thirds,  marked  %  and  so 
on.  These  are  called  duodecimals,  because  it  is  a  regu- 
lar scale,  in  which  the  ratio  between  the  successive  orders 
of  units  is  12  instead  of  10. 

2.  The  addition  and  subtraction  of  duodecimals  are  like 
those  of  other  compound  numbers.     See  LI.  and  LII. 

3.  As  regards  multiplica- 
ation,  let  it  be  proposed  to 
multiply 

The  operation,  performed 
by  the  method  of  aliquot 
parts,  would  be  as  shown 
here. 


The  answer  being  •  go/^.  2     4' 

4.  But  of  what  nature  is  it?  Are  the  60  feet  linear, 
square,  or  cubic?  It  is  what  nothing  in  the  statement 
enables  us  to  determine. 

The  question  might  be,  for  instance, 

1st.  The  trimming  of  a  dress  is  24/j5.  1'  6"  in  length, 
and  each  foot  requires  2ft,  b'  4"  of  ribbon :  how  much 
ribbon  is  wanted  for  the  whole? 

Ans,  QOft,  2'  4',  linear  measure. 

2d.  A  foot-bridge  is  24/^  1'  ^"  long,  and  2ft,  b'  4'' 
wide :  how  many  square  feet  of  plank  are  required  for 
its  floor?  Ans.   60ft,  2'  ^",  square  measure, 

3d.  A  surface  of  ground,  measuring  2^ft,  1'  6'\  square 
measure,  is  to  be  excavated  2ft.  b'  4''  deep :  how  many 
cubic  feet  of  earth  will  be  dug  out  ? 

Ans,  60ft,  2'  4",  cuhic  measure. 

5.  These  examples  show  that  it  is  not  correct  to  affirm 
that  feet  multiplied  by  feet,  or  inches  by  inches,  give 
always  square  feet  or  square  inches  in  the  product.  And 
again,  I  repeat,  let  the  intelligent  pupil  to  whom  such 


DUODECIMALS.  241 

a  multiplication  is  proposed,  ask  for  a  more  precise  state 
ment. 

Whatever  may  be  the  nature  of  the  answer,  the  ope- 
ration may  be  performed  readily  by  aliquot  parts,  as 
above ;  and  there  does  not  appear  to  be  any  necessity  for 
any  additional  rule. 

Arithmeticians,  however,  have  thought  proper  to  in- 
troduce another  method,  which  is  sometimes  used  by 
mechanics  to  measure  the  square  or  solid  feet  of  their 
work.     This  is  the 

MULTIPLICATION  OF  DUODECIMALS. 

6.  Taking  the  same  example  as  above,  the  operation  is 
performed  as  follows  : 

OPERATION. 


Product  by  4" 
«  by  5' 
"      by  2/jf. 


2Aft.  7'    6" 
2ft.  5'    4" 

8' 
10ft.  3 
49      3 

2"     6'"  0"" 
1       6 
0 

60/^.  2'     4"     0"'  0"" 


If  we  consider  that 

1'  =  J  J  of  a  foot. 

V'  =z  J^'  =  ^2  X  yi^  of  a  foot. 

1"'  =  iV  =  T2  of  V2'  =  T2  X  tV  X  A  of  a  foot. 

I..  ^  _Y^  ^  _i^  ^f  _i^//  =  j^  ^f  j^  ^f  ^y  ^  _i^  X  ,V  X 

T2  X  j^^  of  a  foot. 

It  will  be  seen  that  each  additional  index  (')  signifies 
a  division  by  12,  and  that  the  product,  I'X  T  will  be 

1X1    -TgX  12X12--^    • 

^    ^  ^    ~  12X12  X  12X12""  ^     • 
&c.  &c. 

21  Q 


242  LESSON   LVIII. 

7.  That  is,  the  'product  of  two  units  will  give  a  unit 
of  an  order  designated  by  the  sum  of  the  indices  of  the 
two  factors. 

Consequently,  there  will  be  in  the  product  as  many 
duodecimals  as  there  are  in  both  factors  together. 

This  is  a  rule  similar  to  that  observed  in  decimal  frac- 
tions {XXV.,  3),  as  might  have  been  anticipated. 
Therefore, 

I.  In  the  multiplication  of  duodecimals,  multiply  suc' 
cessively  all  the  orders  of  the  multiplicand  by  the  orders 
of  the  multiplier,  beginning  with  the  lowest, 

II.  The  index  of  each  product  will  be  the  sum  of  the 
indices  of  the  factors, 

III.  Set  each  product  in  the  column  of  its  order;  and, 
if  it  exceed  12,  carry  1  to  the  next  superior  order  for 
every  12  it  contains, 

IV.  Continue  to  reduce  and  carry  until  you  reach  the 
highest  denomination, 

V.  The  sum  of  the  several  amounts  will  be  the  answer. 

This  method  is  certainly  a  good  exercise ;  but,  if  the 
two  are  well  understood,  that  by  aliquot  parts  will  be 
found  fully  as  easy  in  practice  and  less  liable  to  error. 

8.  If  there  was  another  factor,  5/>.  IT,  for  example, 
the  duodecimal  multiplication  would  be 

60/lf.      2'     4" 
5ft.    11' 


55/if.      2'     1"     8'" 
300         11      8 

356/)f.      r~~9''    8'" 


But  it  does  not  follow  that  the  product  would  be  solid 
measure  ;  it  might  be  linear,  square,  or  cubic  measure, 
according  to  the  character  of  the  question,  of  which  I 
need  not  give  any  example,  after  the  preceding  explana- 
tions. 

o  v.. 


DUODECIMALS.  243 

I 

9.  In  the  measurement  of  work,  it  is  usual  to  limit  the 
subdivisions  to  inches.  In  that  case,  I  would  advise  the 
use  of  fractions,  and  more  especially  of  decimal  fractions, 
into  which  any  number  of  inches  can  readily  be  trans- 
formed, near  enough  for  practical  purposes. 

Let  it  be  proposed,  for  instance,  to  measure  a  floor 
whose  sides  are  ISft.  6in,  and  14/(f.  3'. 

You  can  change  these  readily  operation. 

into 18.50 

and      .         .         •         •         .  14.25 


9250 
3700 
7400 
1850 


The  product  of  which  is       .        263.6250*^./^. 

The  advantage  of  this  transformation  will  be  still  greater, 
if  the  product  has  to  be  multiplied  by  a  third  factor; 
and  when  you  have,  besides,  to  multiply  the  measurement 
by  the  price.  If  the  price  were  4J  cents  a  square  foot, 
you  would  find  it  easier  to  multiply  263.625  by  0.045, 
than   263ft,  1'  6"  by  4^  cents. 

Questions. — In  measuring  work,  how  is  the  foot  divided  ?  How 
many  seconds  in  a  prime;  thirds  in  a  second?  &c.  What  are 
duodecimals  ?  How  would  you  add  and  subtract  them  ?  Do  feet 
by  feet  always  give  square  feet  ?  When  do  they  not  ?  What 
may  they  give  ?  Give  examples.  Would  the  method  by  aliquot 
parts  answer  to  multiply  duodecimals?  What  is  the  method 
used  generally?  Which  do  you  prefer  ?  What  is  the  rule  of  in- 
dices in  the  multiplication  of  duodecimals?  How  many  duodeci- 
mals will  there  be  in  the  product  ?  Give  the  rule  of  multiplica- 
tion of  duodecimals.  Could  decimals  be  used  in  the  calculations 
of  measurements  of  work?  How?  Would  it  be  as  convenient? 
As  liable  to  error  as  duodecimals  ? 

EXERCISES. 

1.  Multiply  9/)f.  6'  by  Aft.  9'.  Ans.  45/^  1'  6". 

2.  What  is  the  price  of  a  marble  slab,  5ft,  7'  by  1ft.  10',  at 
one  dollar  per  square  foot?  Ayis.  $10.22. 

3.  There  is  a  house  with  three  tiers  of  windows,  three  in  a 
tier:  the  height  of  the  first  tier  is  1ft,  10';  of  the  second,  6/V.  8'j 


244  LESSON   LVIII. 

% 

of  the  third,  5ft,  4';  and  the  breadth  of  each  3/V.  11' :  what  will 
the  glazing  come  to,  at  14  pence  per  foot  ? 

Ans.  £13  llsh.  10|^. 

4.  A  bale  nneasures  1ft,  6'  by  3ft,  3',  and  Ift,  10':  what  are  its 
solid  contents  ?  Ans.  Uft.  8'  3". 

5.  A  merchant  imports  6  bales,  of  the  following  dimensions : 
No. 


Length. 

Height. 

Depth. 

1. 

2ft,  10' 

2ft.  4' 

1ft.  9' 

2. 

2       10 

2       6 

1       3 

3. 

3         6 

2       2 

1       8 

4. 

2       10 

2      8 

1       9 

5. 

2       10 

2       6 

1       9 

6. 

2       11 

2       8 

1       8 

What  is  the  whole  solid  content  and  the  freight,  at  20  cents  per 
foot?  A71S.  Co7ite7it,  '111.64ft, 

Freight,  $14,328. 

6.  How  many  solid  feet  of  timber  in  a  piece  of  timber  2ift, 
6'  long,  \ft.  2'  6"  wide,  and  10'  10"  thick  ? 

A?is.  26C.ft.  8'  8"  6'"  6'"'. 
Decimally,    24.5  X  1.2081  X.903     =26.72  + 

7.  2ft.  Ain.  of  one  kind  of  timber  are  exchanged  for  each  run- 
ning foot  of  another  kind :  how  many  running  feet  must  be  given 
for  ASft.  9171.  ?  Ans.  ll'Sft.  9'  luiearfeet; 

or,  113.75. 

8.  On  each  square  foot  3ft.  5i?i.  of  strips  of  wood  are  used  for 
ornament :  how  many  feet  of  strips  will  be  wanted  for  lOosq.  ft.  6'? 

Ans.  SQOft.  5'  6"  Ihiear  feet; 
or,  360.46,  or  360|i 

N.  B. — Here  is  an  example  of  square  feet  by  linear  feet,  which 
give  linear  feet  in  the  product. 

9.  What  will  be  the  plastering,  at  20  cents  a  yard,  of  a  ceiling 
1yd.  8lfi.  by  4yd.  2ft.  lOin.  Ans.  $7.14. 

10.  What  will  be  the  cost  of  paving  a  surface  19yd.  \ft.  6in. 
by  l%yd.  9i7i.,  at  4|  pence  per  yard  ?  Ans.  £l  Osh.  lOd, 

11.  How  much  wood  in  a  load  1ft.  6'  long,  3ft.  3'  wide,  Ift.  10 
high  ?  A71S.  44ft.  S'  3", 

12.  For  one  solid  foot  of  tinnber  3sq.  ft.  2Q)sq.  in.  of  plank  are 
given :  how  many  running  feet  of  plank,  one  foot  wide,  must  be 
given  for  275  solid  feet  1,448  cubic  inches  ? 

Here  is  a  multiplication  of  solid  by  square  feet,  giving  running 
feet  in  the  answer.  What  answer  would  be  given  if,  without 
further  explanation,  solid  by  square  feet  were  to  be  multiplied  ? 
Here  the  answer  may  be  obtained  by  decimal  fractions,  by  vul- 
gar fractions,  or  by  duodecimals. 


COMPOUND    DIVISION.  245' 

LESSON  LIX. 
COMPOUND  DIVISION. 

1.  Let  us  first  recollect  that  the  dividend  being  the 
product  of  the  divisor  and  quotient,  one  of  these  must  be 
of  the  nature  of  the  dividend,  and  that  the  other  may  be 
anything  else  (XVIII.,  2). 

2.  In  fact,  division  may  also,  like  multiplication,  be 
considered  as  a  species  of  rule  of  three,  in  which  one  of 
the  terms  is  one  unit  of  a  known  nature  ;  so  that  we  have 
either 

The  divisor  :  the  dividend  :  :  1  :  the  quotient. 
Or,   The  divisor  :  1  :  :  the  dividend  :  the  quotient. 

In  the  first  case,  the  divisor  and  dividend  being  of  the 
same  nature,  the  quotient  is  of  the  nature  of  the  known 
unit.     For  example  : 

Cloth  has  been  bought  for  60  dollars,  at  12  dollars  a  yard : 
how  many  yards  have  been  bought  ?     Here  we  have  $12  :  $60  :  : 

lyd,  :  the  answer  =  lyd.  X  rr-^  =  ^j/d. 

The  answer  being  one  yard,  taken  as  many  times  as  the 
ratio  of  $60  to  $12;  that  is,  repeated  as  many  times  as 
the  abstract  quotient,  ^| ;  it  follows  that  the  quotient 
assumes  the  character  of  the  given  unit,  which  is  one 
yard,  though  the  numbers  divided  are  dollars, 

3.  In  the  second  case,  the  divisor  and  known  unit  are 
of  the  same  kind ;  and,  consequently,  the  quotient  com- 
pares with  the  dividend,  and  is  of  the  same  nature. 
Example : 

12  yards  of  cloth  have  been  bought  for  60  dollars :  how  much 
is  it  a  yard  ? 

12y^.  :  hjd.  :  :  $60  :  $60  X  z^~^ 
\2yd. 

Which  shows  that  the  answer  is  $60,  multiplied  by  the 
abstract  quotient  of  1  to  12 ;  or,  in  other  words,  $60 
divided  by  the  abstract  number,  12;  and  the  quotient  is 
of  the  nature  of  the  dividend. 

4.  It  will  appear,  from  what  has  just  been  explained, 
that  before  proceeding  to  divide,  the  nature  of  the  quo- 
tient must  be  inquired  into  from  the  data  of  the  question. 
We  will,  therefore,  distinguish  two  cases. 

21* 


246  LESSON  LIX. 

CASE    I. 

5.  When  the  dividend  and  divisor  are  of  the  same  nature. 

Then  the  quotient  is  of  the  nature  of  the  single  unit, 
referred  to  in  the  question  as  equivalent  to  the  divisor. 

In  this  case, 

L  Change  both  numbers  into  units  of  the  lowest  denO' 
mination  they  contain, 

II.  Divide  the  two  results^  talcing  care  to  convert  suc' 
cessively  the  remainders  into  smaller  denominations  of 
the  kind  indicated  by  the  nature  of  the  question, 

FIRST    EXAMPLE. 
One  yard  of  a  certain  work   is  worth  £23  19*^.  8J^.:  how 
much  can  be  executed  for  £2,728  llsh.  lOd.'i 

It  is  clear  that  the  required  number  of  yards,  multi- 
plied by  ^23  195^.  81J.,  must  produce  ^£2,728  17sh. 
lOd,  Therefore,  the  aggregate  must  be  divided  by  the 
part,  and  the  operation  is  a  Division. 

OPERATION. 
£2728  llsh.  lOd,  £23  19^^.  8J<f. 

20  20 


54577  479 

12  12 


654934  5756 

2  2 


1309868  11513 

11513 


1309868 
15856 
43438 

8899 
3 


U37/d.  2ft.  3m.  9'  iTTfl' 


26697  feet. 
3671 
12 

44052  inches. 
9513 
12 


114156  lines  or  seconds. 
10539 


COiMPOUND    DIVISION.  247 

The  two  numbers,  being  reduced  to  their  lowest  de- 
nomination, become  the  fractions 

■^-f  II---  of  a  pound,  and  mi^  also  of  a  pound ; 
and  we  shall  have  as  many  yards  for  the  answer  as  the 
former  fraction  contains  the  latter;  the  original  denomi- 
nation of  the  numbers  having  nothing  to  do  with  the 
character  of  the  quotient,  which  must  be  in  yards,  though 
pounds  are  apparently  divided.  ^ 

The  operation  is  consequently  reduced  to  a  simple  di- 
vision of  abstract  fractions ;  and,  because  they  have  the 
same  denominator,  we  have  only  to  divide  their  numera- 
tors (XLIV.,  7),  since  it  amounts  to  increasing  both  the 
dividend  and  divisor  480  times,  which  does  not  change 
the  quotient.     Therefore,  the  result  is, 

U/^    V    1309861.     __     130986.8_^,^    . 
iya.  X  -TT5  1  J     ?    O^J      1  1  5  1  3  ~y"'-  • 

which  we  tranform  into  integers,  according  to  the  method 
of  Lesson  LI. 

6.  It  is  clear  that,  in  this  division,  the  nature  of  both 
the  dividend  and  divisor  disappears,  and  only  their  ab- 
stract numerical  ratio  remains,  by  which  the  given  unit 
being  multiplied,  a  neio  denomination  is  substituted, 

I  introduce  these  remarks  here  because  I  have  seen 
pupils  frequently  express  astonishment  that  a  remainder  of 
dollars  or  pounds  (as  they  thought)  should  be  converted 
into  feet,  ounces,  or  other  denominations  than  money. 

SECOND    EXAMPLE. 

One  pound  sterling  has  been  given  for  15lb.  4-02. :  how 
much  will  329Z6.  5oz.  i4.dr.  cost  ? 

OPERATION. 
lb.    oz.  dr,  lb.   oz.  8431813904 

329     5     14  15     4  -G^SS 


16  16  2334|^21  11^^.  lW.-//2» 

20 


1979  244 

329  16  46680  shillings. 

526^  3904  '^^"^^ 


16 


3736 
12 


31628 


^9fiq  44832  pence. 

^^^^  5792 

84318  1888 


248  LESSON   LIX. 

By  changing  here  both  numbers  into  drams,  we  reduce 
the  operation  to  the  division  of 

~IU-  of  a  pound,  by  \\%''  of  a  pound; 

or,  as  in  the  preceding  case,  to  the  abstract  ratio, 

VtoV  multiplied  by  £l,  which  is  at  last  £^¥04^' 

7.  Evidently,  in  this  case,  if  one  of  the  numbers  was 
not  a  compound  number,  it  must  nevertheless  be  reduced 
to  the  same  lowest  denomination  as  the  other,  since  the 
ratio  of  the  dividend  and  divisor  must  be  between  units 
of  the  same  numerical  dimension,  in  order  that  both  may 
be  made  whole  numbers,  by  striking  out  their  equal  deno- 
minators. 

For  example : 

If  one  hundred  weight  of  sugar  costs  £12,  how  much  should 
be  given  for  £55  5sh.  i 

The  dividend  and  divisor  are  both  changed  into  shil- 
lings, and  then  the  result  is  in  hundred  weight,  quar- 
ters, &c. ;  the  value  of  the  fraction,  ^^^-^  X  Icwt.  = 
^cwL  2qr,  llfZ5. 

Questions. — In  division,  of  what  nature  may  the  quotient  be  ? 
When  is  it  of  the  nature  of  the  dividend  ?  What  else  may  it  be  ? 
When  ?  Might  a  division  be  stated  as  a  proportion  ?  How  ? 
How  will  a  proportion  show  the  nature  of  the  quotient  ?  What  in- 
quiry should  precede  the  operation  of  division?  How  many  cases 
are  there  in  compound  division  ?  What  is  the  first  case  ?  What 
is  the  nature  of  the  quotient  in  this  case  ?  What  operation 
should  precede  the  division  ?  Give  the  rule  for  the  operation. 
How  do  you  get  smaller  integers  in  the  quotient  ?  Explain  how 
the  denomination  is  changed  from  one  measure  to  another.  If 
one  of  the  numbers  was  not  compound,  should  it  also  be  reduced  ? 
Why? 

EXERCISES. 

1.  At  £3  155^.  Qd.  per  hundred  weight,  how  much  sugar  can  be 
bought  for  £654  Qsh.  9ff?.  ?  Aiis.  lldcwt.  Iqr.  Ulh, 

2.  At  $14.40  per  hundred  weight,  how  much  can  be  bought  for 
$544.50  ?  Ans.  STcwt.  3qr.  lib. 

3.  At  $18.50  per  acre,  how  much  land  can  be  purchased  for 
$4,252,455  ?  Ans.  229A.  3R.  18P. 


COMPOUND    DIVISION. 


249 


4.  How  many  feet  of  plank,  at  SSsh.  6d.,  per  thousand,  can  be 
bought  for  £iG  llsh.  U.l  Ans.  S,lQ>lft. 

5.  At  $2.25  per  rod,  what   length  of  road  can  be  made  for 
$9,676?  A?is,  13mL  UOR,  2yd,  lift, 

LESSON  LX. 


CASE  II. 

1.  When  the  dividend  and  divisor  are  of  different 
natures,  the  quotient  is  of  the  nature  of  the  dividend. 

In  questions  like  the  following, 

35  yards  of  stuff  have  cost  £51  3sk.  Id, :  what  is  the  price  of 
one  yard  ? 

2.  The  divisor  being  a  whole  iiumbe?* of  the  denomination 
of  the  single  unit  included  in  the  question,  which  in  this  in- 
stance is  the  yard,  it  must  be  considered  as  an  abstract  num- 
ber (LIX.,  3),  since  the  individual  price  of  one  yard  must 
be  3^L  of  the  cost  of  35  yards,  a  ratio  with  which  the 
precise  nature  of  the  articles  compared  has  nothing  to  do. 

The  division  is  here  performed  in  a  manner  similar  to 
simple  division. 

OPERATION. 

£51  3sk,  4d.\S5 
22  \£l  I2sh.  8d, 

20 


You  divide,  first,  the  units 
of  the  highest  order;  the 
quotient  is  =£1,  with  a  re- 
mainder, £22. 

You  change  this  remainder 
into  units  of  the  next  lower 
denomination,  taking  care  to 
add  the  units  of  this  new  de- 
nomination, contained  in  the 
original  dividend;  and  thus 
get  443  shillings. 


440  shillings. 
3 

443 
93 
23 
12 

276  pence. 

4 

"280 


This  new  dividend  of  inferior  units  gives  you,  in  the 
quotient,  the  smaller  integers,  12  shillings,  for  the  second 
part  of  the  quotient,  with  a  remainder,  23  shillings. 

In  the  same  way,  the  second  remainder  is  changed  into 
the  next  lower  order  of  units,  to  which  is  added  the  num- 


250 


LESSON   LX. 


ber  of  similar  units  of  the  original  dividend  :  making, 
here,  280  pence,  which,  divided  by  35,  give,  for  the  last 
integers  of  the  quotient,  8  pence. 

3.  When  the  divisor  also  is  a  compound  number,  the 
operation  must  be  brought  to  the  preceding  case  by  a 
transformation  of  the  divisor,  as  follows : 

I.  Reduce  the  divisor  to  its  lowest  denomination* 

II.  Then  multiply  each  term  of  the  dividend  by  the 
ratio  of  the  lowest  denomination  of  the  divisor,  to  its 
unit  of  comparison, 

III.  Divide  this  altered  dividend  by  the  new  divisoVy 
considered  as  an  abstract  number. 


9cwt.  Iqr,  bib,  have  cost  j63,257  9sh,  Sd,:  how  much 
is  it  per  hundred  weight  ?  The  statement  of  the  opera- 
tion would  be 

^3,257  9sh.  3d.  X  ■,JX;_  ,,,.  (LIX.,  3). 

In  order  to  establish  the  ratio  by  which  we  multiply, 
we  change  both  its  terms  into  pounds ;  that  is,  one  hun- 
dred weight  is  112  pounds,  and  the  divisor  1,041  pounds; 
so  that  the  statement  is  changed  into 

^3/257  9sh.  3d.  X  j^^j. 

And  all  we  have  to  do  is  to  multiply  each  part  of 
the  dividend  by  112,  and  then  divide  by  1,041.     Thus: 


4.  This  manner  of 
proceeding,  it  willbe 
perceived,  amounts 
to  reducing  the  di- 
visor to  its  lower 
denomination, 
1041/^.  =  \\V  of 
one  hundred  weight. 


OPERATION 

£3,257 

Qsh. 

3d. 

112 

112 
lOOSsh. 

112 

£364784 

336d. 

5248 

•434 

20 

shillings 

9688 

. 

319 

12 

1041 


£350  9sh,  4d. 


4164  pence. 


COMPOUND   DIVISION.  251 

And  then  making  it  a  whole  number,  as  usual  in'  divi- 
sion by  fractions,  by  striking  out  its  denominator  ;  which 
operation  rendering  it  112  times  larger,  we  must  also 
multiply  the  dividend  by  112,  that  the  quotient  may  not 
be  altered. 

5.  It  will  be  remarked  that,  in  multiplying  by  112,  I 
do  not  reduce  the  lower  products  and  carry  to  the  higher 
denominations.  This,  though  generally  done,  is  lost 
labor;  since  the  division  of  the  successive  products  will 
generally  give  small  quotients,  as  in  this  example. 

But,  even  if,  in  some  rare  case,  the  number  of  units  of 
some  of  the  lower  denominations  should  exceed  the  value 
of  an  integer  of  the  next  higher  order,  the  reduction 
would  always  be  readil}^  made  afterwards. 

If,  for  instance,  the  dividend  were  .£146  16sh.  3d,, 
and  the  divisor  Icwt,  Iqr.  5lb.,  the  quotient,  by  our 
method,  would  be 

£112  275^  12d.] 

which  would,  without  difficulty  or  hesitation,  be  changed 
at  once  into 

£113  8s^. 

6.  N.  B. — In  such  questions,  be  careful  to  observe 
which  of  the  units  of  the  divisor  is  the  unit  of  compari- 
son, whose  value  is  required  3  because  it  is  in  regard  to 
this  unit  that  the  divisor  is  to  be  transformed. 

For  instance,  if,  in  the  first  question,  it  was  a  foot  of 
stuff,  the  value  of  which  were  required,  the  35  yards 
must  be  changed  into  105  feet. 

If,  in  the  second,  it  was  the  ton  whose  value  were  re- 
quired, the  denominator  of  the  transformed  divisor  should 
be  2,240,  instead  of  112. 

If  it  were  the  quarter,  then  the  denominator  should 
be  28. 

If  the  'pound,  the  reduction  to  the  lowest  denomination 
would  be  the  only  preparatory  operation  necessary. 

Questions. — What  is  the  second  case  in  compound  division  ? 
How  do  you  divide,  if  the  divisor  is  a  whoJe  number  ?    How, 


Ans.  i:20  135/2. 

,Sd, 

Ans.  £         sh. 

d. 

Ans.  £         sh. 

d. 

lb.    oz.    dtvt. 

gr. 

ns,       cwt,    qr. 

lb. 

252  LESSON   LX. 

when*  it  is  a  compoimd  nnmber?  By  what  number  should  the 
terms  of  the  dividend  be  multiplied  ?  Why  ?  Explain  it  by  the 
arithmetical  statement.  Explain  it  by  making  the  divisor  a 
whole  number.  Is  it  necessary  or  proper  to  reduce  the  indi- 
vidual products  of  the  dividend?  What  attention  should  be  had 
in  regard  to  the  unit  of  comparison  ?     Give  examples. 

EXERCISES. 

1.  Divide  £22^  lOsk.  4d.  by  11. 

2.  «         jesi  2sh.  lOld.  by  99. 

3.  «         £315  3sh.  lO^d.  by  135. 

4.  «         23/^.  loz.  Qdwt.  12gr.  by  7. 

A71S. 

5.  "         IjOeicwt.  2qr.  by  28.  A 

6.  «         315ml,  2fur.  IP,  2yd.  \ft.  2in.  by  39. 

Ans.  9ml.  Aftir.  39P.  2ft.  8|m. 

7.  If  72  bushels  cost  £20  9*^.  6J.,  what  is  it  per  bushel  ? 

Ans.     sh.      d. 

8.  If  1  hundred  weight  cost  23^^.  4J.,  what  is  it  per  pound  ? 

Ans.  2\d, 

9.  When  2  hundred  weight  of  sugar  cost  £8  17^^.  4(^.,  what 
is  it  per  pound  ?  Ans.  9\d, 

10.  If  an  ingot  of  gold,  weighing  9/5.  9oz.  12dwt.y  be  worth 
£411  125/i.,  what  is  that  per  grain  ?  Ans.  l^d, 

11.  A  person  breaking,  owes  £1,490  5sh.  10^.,  and  has  £784 
Xlsh.  Ad, :  what  will  each  creditor  get  on  the  pound  ? 

Ans.  lOsh,  6i^. +|ff|-f. 

12.  If  42cwt.  Iqr,  Ulb,  cost  $63.56^,  what  is  it  per  hundred 
weight?  A71S.  $1.50. 

13.  If  16cwt.  2qr.  lllb.  of  sugar  were  sold  for  £46  Ush.  Id., 
how  much  is  that  per  hundred  weight  ?  Ans,  £2  I5sh.  2d. 

14.  If  $17,113,637  have  been  paid  for  578^.  3R.,  how  much  is 
it  per  acre  ?  Ans.  $29.57. 

15.  If  69/2  yards  have  cost  £2,728  17*^.  9d.,  how  much  is  it  a 
yard?  A71S,  £39  Ash.  ^y. 

16.  If  y  of  an  ounce  cost  £\i,  what  will  1  ounce  cost  ? 

Ans.  £1  5sh.  8d. 

17.  A  printer  uses  1  sheet  of  paper  for  every  16  pages  of  an 
8vo.  book :  how  much  paper  will  be  necessary  to  print  500  copies 
of  a  book  containing  336  pages,  allowing  2  quires  of  waste  paper 
in  each  ream?  Right  Answer.  24  reams  6  quires  2f  sheets. 


COMPOUND   PROPORTION.  253 


CHAPTER  VIII. 

CONTAINING    PRACTICAL    QUESTIONS    WHICH   DEPEND    ON 
PROPORTIONS. 

LESSON  LXI. 

1.  We  have  explained,  in  Lesson  XLVL,  how  questions 
depending  on  simple  proportion  should  be  stated,  and 
how  the  rule  of  three  is  reduced  to  plain  multiplication 
and  division.  It  may  have  been  remarked  that  the  an- 
swer is  always  of  the  denomination  of  one  of  the  given 
numbers,  and  is  obtained  by  changing  this  number,  ac- 
cording to  the  ratio  of  two  other  numbers. 

2.  It  frequently  happens,  however,  that  practical  ques- 
tions are  more  complicated,  and  that  the  answer  depends 
on  the  known  number  of  the  same  denomination,  by 
more  than  one  condition,  and  that  the  combination  of 
several  ratios  is  necessary.  Such  questions  we  will  now 
proceed  to  examine.  They  may  all  be  classed  under  the 
head  of 

COMPOUND  PROPORTION,  OR  DOUBLE  RULE  OF 
THREE. 

3.  The  following  is  a  simple  example  of  questions  of 
this  sort : 

Example  1st. — 20  men  have  built  50  feet  of  wall  in  18  days : 
how  many  men  will  build  120  feet  of  the  same  kind  of  work  in 
12  days  ? 

All  such  questions  may  be  performed  in  two  different 
ways:  either  by  the  method  of  ratios^  or. by  the  method 
of  units, 

METHOD   OF   RATIOS. 

I.  Write  down  the  number  of  the  same  denomination 
as  the  answer, 

II.  Multiply  it  successively  by  the  direct  or  inverse 
u2i 


254 


LESSON    LXr. 


ratio,  as  the  case  may  be,  of  each  pair  of  given  numhers 
of  the  same  Mnd  ;  the  final  product  will  he  the  answer. 

Observe  that  both  terms  of  a  ratio,  though  of  the  same 
kind,  may  not  be  of  the  same  denomination.  When  this 
is  the  case,  they  should  be  reduced  to  one  and  the  same 
denomination. 

In  establishing  a  ratio,  consider  whether  the  answer 
should  be  increased  or  diminished  by  it.  In  the  first 
case,  place  the  larger  number  in  the  numerator  j  in  the 
second,  place  it  in  the  denominator. 

In  the  question  before  us,  for  instance ;  since  a  number 
of  men  is  required,  we  have  to  determine  how  the  given 
number  of  men,  20,  is  to  be  modified  by  the  conditions 
of  the  question,  so  as  to  be  changed  into  the  answer. 

Beginning  with  the  modification  produced  by  the  num- 
ber of  days,  we  remark  that  it  would  take  a  greater 
number  of  men  to  execute  the  same  work  in  a  less  num- 
ber of  days.  Hence,  the  ratio  is  inverse,  and  we  multi- 
ply 20  by  f|. 

But  now,  since  the  work  is  different  and  greater  in 
quantity,  the  number  of  men  must  be  increased  in  the 
direct  ratio  of  the  quantity  of  work ;  that  is,  in  the  ratio 
of  1/^0^  Hence,  the  final  answer  is  20  X  -j-f  X  VV^  = 
72  men. 

METHOD    OF    UNITS. 

4.  Find  the  answer  for  one  unit  of  each  denominatiouy 
and  then  that  for  a  number  of  each  kind  will  easily  be 
obtained,  either  by  multiplication  or  division,  as  the 
nature  of  the  question  inay  require. 

In  the  above  example,  we  must  first  find  how  many 
men  it  would  take  to  build  one  foot  of  wall  in  one  day. 

Evidently,  if  it  take  20  men  to  build  a  wall  in  18  days, 
it  will  take  18  times  as  many  to  build  the  same  in  one  day; 
that  is, 

20  X  18. 

This  being  the  nmnber  necessary  to  build  50  feet  in 
one  day,  it  will  require  only  the  fiftieth  part  of  it  to 
build  one  foot  j  that  is, 


COMPOUND    PROPORTION.  255 

20  X  18 


50 


-> 


which  is  the  number  of  men  requisite  to  build  one  foot  in 
one  day. 

And  now,  passing  on  to  the  second  series  of  numbers, 
it  is  clear  that  the  number  of  men  necessary  to  build  one 
foot  of  wall  in  12  days,  instead  of  one  day,  will  be  -^^  of 
this  J  that  is, 

20  X  18 

50  X  12' 

and  that  it  will  take  120  times  this  new  number  to  build 
120  feet,  instead  of  one  foot ;  or,  finally, 

20  X  18  X  120        ^^  , 

— ^Q      ^^ —  =  72  men,  as  above. 

5.  The  same  might  be  obtained  by  a  series  of  propor- 
tions multiplied  into  each  other;  but  the  preceding 
methods  are  preferable,  and  proportions  are  no  longer 
used  in  such  questions. 

The  following  more  complicated  example  would  be  as 
readily  calculated : 

Example  2d. — If  9  laborers,  working  8  hours  a  day,  have  spent 
24  days  in  digging  a  ditch  54  yards  long,  14  wide,  and  5  feet  deep, 
how  many  days  will  it  take  72  laborers,  working  11  hours  a  day, 
to  dig  a  ditch  275  yards  long,  18  broad,  and  7  feet  deep  ? 

FIRST    METHOD. 

Here,  a  number  of  days  being  required,  we  multiply 
the  known  number  of  days,  24,  as  follows : 


24  days  by  ^%  by  ^\  by  %^  by  if  by  J  =  20  days. 

The  answer  is  readily  obtained  by  cancelling  9  and  8,  with  72 ; 
11  and  5  out  of  275 ;  18  out  54  j  7  out  of  14;  and,  finally,  3  and 
2  out  of  24. 


250  LESSON    LXI. 

SECOND    METHOD. 


One  laborer,  working  one  hovr  a  day,  would  dig  a 
ditch  one  yard  long,  one  yard  wide,  and  one  foot  deep,  in 

24  X  9  X  8 

54  X  14  X  5  ^^^"^^ 

and,  consequently,  72  laborers,  working  11  hours  a  day, 
would  dig  a  ditch  275  yards  long,  18  broad,  and  7  feet 
deep,  in 

24  X  9  X  8         275  X  18  X  7         ^^   ,  , 

54X14X5  ^        72X11~  =  ^^  ^^y''  ^'  ^^^^'- 

Remark  that,  when  you  have  obtained  the  amount  for  all  the 
units  of  the  data,  the  corresponding  numbers  of  each  kind  are 
evidently  on  the  opposite  side  of  the  line  of  division,  and  may 
be  written  at  once.  Thus,  9  laborers  of  the  first  series  being  in 
the  numerator,  the  corresponding  number,  72  laborers,  is  in  the 
denominator  of  the  second  series  :  8  hours  of  the  first  is  in  the 
numerator;  and  consequently  11  of  the  second  in  the  denomina- 
tor. 54  yards  of  the  first  is  in  the  denominator ;  and,  therefore, 
275  yards  of  the  second  must  be  in  the  numerator,  &c. 

Questions. — What  is  compound  proportion?  By  how  many 
methods  can  questions  in  compound  proportion  be  solved?  Ex- 
plain the  method  by  ratios.  What  is  done  when  two  numbers  of 
the  same  nature  are  not  of  the  same  denomination  ?  How  do 
you  determine  which  of  the  numbers  of  a  ratio  is  to  multiply, 
and  which  to  divide  ?  Explain  the  methods  by  units.  In  this 
method,  after  having  found  the  answer  for  o77.e  of  each  denomina- 
tion, how  should  each  one  of  the  corresponding  numbers  of  the 
second  series  be  placed  ? 

EXERCISES. 

1.  15  men,  working  10  hours  a  day,  have  taken  18  days  to  con- 
struct 450  yards  of  a  certain  work :  how  many  men,  working 
12  hours  a  day,  would  make  480  yards  in  8  days  ?     Aiis,  30  men, 

2.  1,200  yards  of  cloth,  f  of  a  yard  wide,  will  clothe  500  men : 
how  many  yards,  |  wide,  are  necessary  for  960   men  ? 

Ans.  3,291^  yards. 

Here  multiply  1,200  by  the  ratio  of  the  widths  and  by  that  of 
the  men. 

3.  A  man,  walking  15  hours  a  day,  has  travelled  375  miles  in 
20  days  :  how  many  hours  a  day  must  he  travel  to  go  400  miles 
in  18  days  ?  Ans,  177  hours. 


FELLOWSHIP.  257 

4.  If  a  family  of  9  persons  spend  450  dollars  in  5  months,  how 
much  would  be  sufficient  to  maintain  them  8  months,  if  5  more 
were  added  to  the  family  ?  A?is.  $],120. 

Eitner  multiply  $450  by  the  ratios,  or  find  first  what  will  sup- 
port one  person  one  month,  and,  from  that,  what  will  support  14 
persons  8  months. 

5.  If  100  dollars  gain  5  dollars  in  12  months,  how  much  will 
750  dollars  gain  in  7  months?  A71S.  $21,875. 

6.  If  120  bushels  of  corn  can  serve  14  horses  56  days,  how 
many  days  will  94  bushels  serve  6  horses?        A71S.  1021 1  days. 

7.  If  loz,  5dr.  of  bread  can  be  bought  at  4|  pence,  when 
wheat  is  at  45/i.  2d.  per  bushel,  what  weight  may  be  bought  for 
Ish.  2d.,  when  the  price  of  wheat  is  5sh.  Qd.  per  bushel  ? 

Ans.  lib.  5dr.    ^id^at, 

8.  If  the  transportation  of  \3cwt.  Iqr.,  a  distance  of  72  miles, 
cost  £2  lOsk.  6d.j  what  will  be  the  transportation  of  Icwt.  3qr, 
on  112  miles?  A7is.  £2  5sh.  Ud.  IT^  far. 


LESSON  LXII. 

N.  B, — Those  who  intend  learning  Algebra,  had  better  begin  it  now. 
FELLOWSHIP  OR  PARTNERSHIP. 

1.  The  preceding  lesson  completes  all  that  is  strictly 
necessary  to  solve,  with  proper  care  and  reflection,  all 
questions  depending  on  proportion. 

The  following  rules,  however,  being  of  great  practical 
utility,  will  be  noticed  separately,  and  at  some  length, 
more  on  account  of  their  frequent  applications  in  busi- 
ness transactions  than  of  any  real  difficulty  they  present. 

2.  Fellowship  is  a  rule  by  which  a  given  amount  is 
divided  into  a  number  of  parts,  in  a  certain  proportion. 

By  this  rule,  which  is  an  application  of  compound  pro- 
portion, are  adjusted  the  gains,  losses,  and  charges  of 
partners  in  company ;  the  effects  of  bankrupts  j  the  shares 
of  prizes,  &c. 

3.  Fellowship  may  be  single  or  double.  It  is  single 
when  the  shares  or  dividends  depend  only  on  one  condi- 
tion, and  are  consequently  proportioned  to  one  number 
only. 

22*  R 


258 


LESSON   LXII. 


It  is  double  when  each  share  depends  on  several  con- 
ditions 3   it  is  then  a  case  of  compound  proportion. 

SINGLE    FELLOWSHIP. 

4.  Example  1st. — Three  merchants  have  formed  a  copartner- 
ship: A  put  in  $15,000;  ^,$22,540;  C,  $25,600.  At  the  end 
of  a  year  their  joint  profit  amounts  to  $12,000  :  what  is  the  share 
of  each? 

It  is  evidently  conformable  to  justice  to  divide  the 
proceeds  in  proportion  to  the  sum  adventured  by  each. 
Upon  this  principle,  the  process  of  distribution  is  very 
simple. 

Find  first  what  is  the  profit  made  on  each  dollar,  and 
multiply  it  by  the  stock  in  trade  of  each  person. 

Now,  it  is  evident  that  the  profit  on  one  dollar  is  equal 
to  the  whole  profit  divided  by  the  aggregate  capital; 
that  is,  to 

jJt  1  2  0  0  0 

Consequently,  the  statement  of  the  successive  shares 
will  be, 

1st  share,  .         '-^^^^'  =  $2,850.81. 

2d  share,  .        '^>^  =.$4>,283M. 

3d  share,  .        12000X25600  ^  $4^865.38. 

63140  

Being  in  all,  to  prove  the  work,     .       $12,000.00. 

In  business,  the  money  advanced  is  called  the  capital, 
or  stock,  and  the  sum  to  be  distributed  among  the  part- 
ners, the  dividend* 

Example  2d. — A  ship,  vsrorth  $4,800,  is  lost  ;•  of  vrhich  |  be- 
longs to  A,  §  to  Bj  and  the  balance  to  C :  what  is  the  loss  of 
each? 

We  must  first  find  the  proportional  share  of  C,  which 

24* 

Then  we  have  to  divide  the  amount  in  the  proportion 
of  the  three  shares,  ^^j  5^5  and  ^^ ;  that  is,  of  the 
numbers      .         .         3,     8,  13. 


FELLOWSHIP.  259 

Now,  it  Is  evident  that  each  unit  loses  a  part  of  the 
whole  amount,  equal  to  §4,800,  divided  by  the  sum,  24<, 


$200 ; 

$600 
$1,600 
$2,600 

$4,800 


of  the  three  numbers ;  that  is,  to 

and,  consequently,  the  loss  for    3  units  is 

for    8       " 

for  13      " 

Verification, 

DOUBLE   FELLOWSHIP. 

5.  Example  \st. — A  and  B  trading  together,  A  has  in  the 
"business  $500  for  4  naonths,  and  B  $600  for  5  naonths ;  the  pro- 
fits are  $240 :  how  are  they  to  be  divided  ? 

It  is  clear  that  each  dollar  has  produced  a  certain  pro- 
fit in  each  month,  which,  multiplied  by  the  number  of 
months,  will  give  its  whole  profit  while  in  the  business. 

Hence,  it  is  just  that  the  division  should  be  in  propor- 
tion, not  only  of  each  capital,  but  likewise  of  the  time  j 
that  is,  of  the  compound  ratio  of 

500  X  4  to  600  X  5,  or  of  2,000  to  3,000. 

6.  In  fact,  it  is  easy  to  understand  that  $2,000  in  one 
month  will  yield  as  much  profit  as  500  in  4  months,  and 
3,000  as  much  as  600  in  5  months. 

By  thus  multiplying  the  various  proportional  condi-^ 
tionsy  the  compound  ratio  is  reduced  to  a  simple  one^  and 
the  rest  of  the  operation  is  as  in  single  fellowship. 

In  this  case,  the  ratio  being  as  2  to  3,  the  result  is, 

For  the  first,  .         .         $96. 

For  the  second,  .         .       $144. 

Example  2d. — Three  persons  hold  a  pasture  in  common,  for 
which  they  are  to  pay  £30  per  annum.  A  puts  in  7  oxen  for 
3  months ;  J9,  9  oxen  for  5  months ;  C,  4  oxen  for  12  months : 
what  part  of  the  rent  must  each  person  pay  ? 

Each  ox,  of  course,  is  supposed  to  consume  an  equal 
quantity  per  month  ;  and  it  is  evident  that 

7  oxen  consume  as  much  in    3  months  as   7x3  in  one. 

9     "         "        as  much  in    5  months  as   9x5  in  one. 

and  4      '         "        as  much  in  12  months  as  1 2  X  4  in  one. 


260 


LESSON    LXII. 


The  question  is  thus  reduced  to  single  fellowship;  and 
the  division  made  in  the  proportion  of  the  numbers  21, 
45,  48,  gives,  .         for  A,     .       £5     lOsh.    6f-d. 

for  B,     .        11       6       10j% 
for   C,     .        12     12         7-ii 

7.  Questions  like  the  following,  though  differing  in 
some  particular,  may  be  ranked  with  fellowship : 

A  man,  with  the  force  he  employs,  may  complete  a  certain 
work  in  2^  months ;  another  could  do  it  in  3f  months :  if  they 
form  a  partnership,  when  can  they  finish  the  work  ? 

To  solve  this  question,  let  us  find  what  part  of  the 
work  each  will  accomplish  in  one  month. 

The  first  will  make,  in  one  month,  r^  =   |  of  the  work. 
The  second,         ''  «  i  =  /,-. 

Hence,  both  together  will  make  f  +  j\  =  \l  of  the 
same  in  one  month ;  that  is,  -^j  in  yL.  part  of  a  month ; 
and,  consequently,  the  whole  in  -}|  of  a  month,  or  IJ 
months. 

8.  The  same  may  also  be  understood  in  this  way: 
Multiply  the  two  times  together ;  it  is  evident  that,  in  the 
time  2J  X  8|,  the  first  would  do  the  work  3|  times,  and 
the  second  2|-  times. 

Hence  together,  in  this  multiple  time,  they  would  do 
it  2|-  +  3|  times )  therefore,  they  will  do  it  once  in 

23.  .33  =  U  months. 

Questions. — What  is  fellowship ?  Single?  Double?  Dividend? 
How  are  the  shares  found  in  single  fellowship  ?  How,  in  case  of 
fractional  proportions  ?  How  do  you  proceed  in  double  fellow- 
ship ?     What  other  questions  may  be  classed  with  fellowship  ? 

EXERCISES. 

1.  A  ship's  company  take  a  prize  of  i^lOOO,  which  is  to  be 
divided  among  them, •according  to  their  pay  and  the  time  they 


FELLOWSHIP.  261 

have  been  on  board.  The  officers  and  midshipmen  have  been  on 
board  6  months,  the  sailors  3  months.  The  officers  receive  40 
shillings  a  month,  midshipmen  30  shillings,  sailors  22  shillings. 
There  are  4  officers,  12  midshipmen,  110  sailors :  what  is  the  share 
of  each  ?  A7is.    Each  officer,  .     .     £23     "Ish.  5  j^/g  rf. 

"     Tnidshi'pman,     17     6        9i47 
«     sailor,  .     .  6     7        2_|^ 

2.  1,200  horses  are  to  be  distributed  to  three  regiments,  in 
proportion  to  their  numerical  strength.  The  number  of  men  in 
the  first  is  to  that  in  the  second  as  11  to  8,  and  that  of  the  first  to 
the  third  as  9  to  7  :  how  many  horses  must  each  regiment  re- 
ceive? Ans.    \st,     .     .     479  1. 

2^;,      .     .     348i.|. 

3^,      .     .     3724._8. 

'  3  1* 


1,200    0. 

In  a  case  like  this,  which  occurs  frequently,  when  the  distri- 
tribution  does  not  allow  of  fractions,  the  unit  is  allowed  to  the 
largest  fraction.  The  surplus  horse  would  therefore  go  to  the 
third  regiment. 

3.  A  bankrupt  is  indebted  to  A,  $277.33;  to  J5,  305.17;  to  C, 
$152  ;  and  to  B,  $105  :  his  estate  being  worth  $677.50,  how  must 
it  be  divided?  Ans,  {Omitting  fractions^     A,     $223.81. 

jB,  $246.28. 
C,  $122.67. 
Jy,      $84.74. 

4.  An  individual  engages  in  a  certain  enterprise,  with  $25,000; 
five  months  afterwards  he  takes  in  a  partner,  with  $40,000  ;  then, 
after  six  more  months,  he  takes  a  second  partner,  with  $60,000. 
After  two  years,  the  profits  amount  to  $80,000.  It  has  moreover 
been  stipulated,  that  the  first  individual  is  to  be  allowed  5  per 
cent,  on  the  profits,  for  his  management  of  the  concern :  what  is 
the  share  of  each  ?  Ans.    \st,     .     $25,308.41. 

2^/,      .     $26,990.65. 
3^,      .     $27,700.94. 

5.  A  can  mow  5  acres  in  3  days;  B  can  mow  7  acres  in  4  days; 
and  C,  9  acres  in  5  days :  how  long  will  it  require  them,  working 
together,  to  mow  15  acres  ?  Ans,  2.875. 

Find  what  each  can  do  in  one  day,  and  solve  by  No.  7. 
Or,  find  in  what  time  each  will  mow  15  acres,*  and  solve  by 
No.  8. 

6.  A  mound  of  earth,  containing  60  cubic  yards,  can  be  re- 
moved with  a  cart  in  3  days,  of  10  hours  work;  with  a  two-horse 


262  LESSON   LXIII. 

wagon,  in  two  days;  and,  with  a  four-horse  wagon,  in  one  day: 
in  what  time  will  the  three  together  do  it  ?  A?is.  5hr.  5^jm, 

7.  A  reservoir  has  two  cocks  to  supply  it;  by  the  first,  it  may 
be  filled  in  40  minutes;  by  the  second,  in  50.  It  has  also  a  dis- 
charging cock,  by  which  it  may  be  emptied  in  25  minutes. 
These  three  cocks  being  open  together,  in  what  time  would  the 
cistern  be  filled  ?  A7is»  3hr»  20m, 


LESSON  LXIII. 
ALLIGATION. 

1.  This  rule,  which  is  closely  allied  to  fellowship, 
serves  to  find  the  mean  value  or  average  of  several  things. 

Example  \st. — A  wine  merchant  mixes  together  25  bottles  of 
wine,  at  12  cents;  18,  at  15  cents;  and  20,  at  16  cents:  what  is 
the  value  of  one  bottle  of  the  mixture  ? 

It  will  readily  occur,  that  the      25  bottles  cost  |3.00. 
"  18         "  2.70. 

"  20         "  3.20. 

Which  gives,  for  the  aggregate    

value  of  ....  63  bottles,  .  $8.90. 
And,  consequently,  for  one  bottle,  $^^|°  =  \Acts.  Im. 

Example  2d. — 500  men  are  employed,  of  whom  160  are  paid 
$2  a  day;  200,  $1.75;  and  140,  $1.50:  what  is  the  average  price 
of  this  labor  i  Ans.  $1.76. 

2.  This  operation  is  called  alligation  medial. 

The  average  is  obtained  by  multiplying  the  individual 
values  by  the  respective  quantities,  and  dividing  the 
sum  of  the  products  by  the  aggregate  quantity. 

3.  The  reverse,  by  which  the  mean  value  or  average 
being  given,  the  proportion  of  the  component  parts  is 
obtained,  is  distinguished  by  the  name  of  alligation  alter^ 
nate. 

Example  \st. — How  much  wine,  at  $2  per  gallon,  should  be 
mixed  with  wine  at  $3,  so  as  to  make  a  mixture  worth  $2.75  per 
gallon  ? 

It  is  clear,  that  each  gallon  of  the  first  wine  will  gain 


ALLIGATION.  263 

75  cents  in  the  mixture,  and  each  gallon  of  the  second 
will  lose  25  cents )  and  also,  that  the  quantities  mixed 
must  be  such  that  the  gain  by  the  one  may  be  ^qual  to 
the  loss  by  the  other  ;  a  result  which  will  be  accom- 
plished by  taking  quantities  in  reverse  proportion  of  the 
loss  and  gain;  that  is,  in  the  proportion  of  25  to  75; 
since  the  products  will  be  equal ;  that  is,  in  this  case, 
the  gain  75  cents  X  25  =  25  cents  X  75  the  loss. 

So  that  0.25  or  |  of  a  gallon  of  the  dearer  wine,  and 
0.75  or  I  of  the  cheaper  will  make  one  gallon 
of  the  mixture. 

4.  This  process  of  reasoning  proves  that  the  proportion 
of  the  component  parts  is  found  by  taking  the  difference 
between  the 'price  of  each  and  that  of  the  compound:  the 
quantities  are  in  an  inverse  ratio  of  these  differences, 

5.  When  the  quantity  of  the  compound^  besides^  is 
fixed y  each  component  will  be  obtained  by  multiplying 
the  quantify  of  the  compound  by  the  proportional  quan^ 
tity  of  each  component  part. 

Thus,  if  60  gallons  of  wine  were  to  be  made  by  the 
above  conditions,         .         ^  of  60  =  15, 
and     .         .         .        1  of  60  ==  45, 
would  be  the  respective  quantities. 

6.  When  the  quantity  of  one  of  the  parts  is  given  to^ 
gether  with  all  the  prices : 

I.  After  having  found  the  proportion  of  the  compo^ 
nentSy  get  the  second  part  by  multiplying  the  quantity  of 
the  first  by  the  ratio  between  the  two, 

n.   Their  sum  will  be  the  quantity  of  the  compound. 

Example. — How  many  gallons  of  water  must  be  mixed  with 
SO  gallons  of  wine,  at  $6  per  gallon,  to  make  a  mixture  worth  $5 
per  gallon? 

In  the  first  place,  we  find  the  ratio  between  the  quan- 
tities to  be  as  5  :  1. 

Therefore,  we  must  add  ^  of  water,  or  6  gallons. 

7.  Alligation  alternate  may  sometimes  be  proposed  be- 
tween a  greater  number  of  ingredients ;  as,  for  example, 


264 


LESSON   LXII. 


between  tea,  at  $1,  at  $1.50,  and  at  75  cents,  to  be  worth 
$1.25. 

It  is  very  clear,  that  the  mixture  may  be  made  in  a 
great  variety  of  ways;  since,  if  we  combine  two  of  them 
in  any  way  (provided  only  that  the  general  mixture  may 
still  remain  between  the  partial  mixture  and  the  remain- 
ing ingredient),  these  two  may  be  mixed  by  the  regular 
rule. 

The  operation  would  be  more  indefinite  and  arbitrary 
still,  if  the  number  of  ingredients  were  increased  ;  in 
fact,  such  a  problem  is  of  no  importance  in  practice. 

Questions. — What  is  alligation?  Medial?  Alternate?  How 
is  the  average  found  when  the  quantities  and  values  are  given  ? 
How,  the  proportion  of  components,  when  both  prices  and  their 
average  are  given  ?  How,  the  two  quantities,  when,  besides 
prices,  the  quantity  of  the  compound  is  given  ?  How,  the  second 
part  and  the  aggregate,  when  prices  and  the  quantity  of  one  of 
the  components  are  known?  How  is  alligation  alternate,  in  case 
of  several  compounds  ? 

EXERCISES. 

1.  A  goldsmith  mixes  8M.  ^\oz.  of  gold,  of  14  carats  fine,  with 
X^lh.  8^02:.,  of  18  carats:  what  is  the  fineness  of  the  mixture? 

Ans,   l-G  5  1     carats, 

2.  If  I  mix  27  bushels  of  wheat,  at  ^slu  6r/.  the  bushel,  with 
the  same  quantity  of  rye,  at  \sh.  per  bushel,  and  14  bushels  of 
barley,  at  2sh.  Sd.  per  bushel,  what  is  the  worth  of  a  bushel  of 
the  mixture  ?  A7is,  Ash.  3|  + 1_3^ 

3.  A  contractor  wishes  to  bid  for  20  miles  of  road ;  he  esti- 
mates 3  miles  at  $875  per  mile ;  5,  at  $1,500 ;  6,  at  $650 ;  4,  at 
$1,100 ;  and  2,  at  $2,000  :  what  must  be  his  average  bid  per  mile  ? 

A71S.  $1,121.25. 

4.  In  grading  a  road  up  a  mountain,  by  an  experimental  survey, 
it  is  found  that  the  rise  of  45  rods  is  at  lO^^ ;  22  rods,  at  9|^ ;  38, 
at  9°;  60,  at  5^;  110,  at  4°;  and  55,  at  S'^ :  what  should  be  the 
uniform  grade  of  the  road,  allowing  a  level  platform  of  2  rods  in 
every  60  rods?  Ans,  5°  51%', 

5.  A  survey  is  run  through  woods ;  the  first  course  is  120  poles 
long;  the  second,  85,  and  deviates  3*^  to  the  right  of  the  first ;  the 
third,  95  poles,  and  deflects  from  the  first  5^,  also  to  the  right; 
the  fourth,  125  poles,  and  deflects  6^"  to  the  left  of  the  first ;  the 
fifth,  75  poles,  deflecting  1«^  to  the  right  j  finally,  the  sixth,  50 


PERCENTAGE.  265 

poles,  deflecting  2°  to  the  left :  what  is  the  straight  direction  be- 
tween both  endsj  in  regard  to  the  first  course  ? 

A71S.  0°  4yy'  to  the  left  of  first  cotirse. 

Multiply  distances  by  grades,  and  subtract  between  the  pro- 
ducts of  the  left  and  right ;  then  divide  by  the  whole  distance  : 
the  result  will  show  the  deflection  on  the  side  of  the  larger  pro- 
duct. 

N.  B. — The  two  preceding  questions  are  convenient  and  cor- 
rect in  practice,  for  angles  not  exceeding  12^^  or  15^. 

6.  On  a  railroad  there  were  carried  15,750  tons  173  miles ; 
6,820  tons,  82  miles;  9,719  tons,  32  miles.  The  cost  of  repairs 
of  cars  has  been  $17,188.60  :  what  is  it  per  ton  per  mile  ? 

A71S.  0.478  ct, 

7.  A  manufacturer  wants  to  make,  with  a  mixture  of  cotton, 
linen  that  will  sell  at  $1  per  yard,  while  pure  linen  would  sell  at 
$1.50,  and  cotton  stuff  at  25  cents,  all  of  the  same  fineness  :  what 
proportion  of  cotton  and  flax  must  be  used  ? 

Ans,  2  of  cotton  to  3  of  flax, 

8.  A  man  can  finish  a  piece  of  work  in  30  days,  with  240  men, 
working  10  hours  a  day.  He  engages  100  men,  who  agree  to 
work  12  hours ;  but  all  those  who  are  yet  for  hire,  consent  to 
w^ork  only  9  hours  :  how  many  of  these  must  be  engaged  to  finish 
in  time?  Ans.   133  men  {omitting  the  fraction  ^). 

9.  Make  120  gallons  worth  6.s/i.  per  gallon,  with  wines  worth 
^sh.  and  8jA.  Ans.  80  at  5sh, 

40  at  Ssh. 

10.  A  grocer  would  mix  teas,  worth  8  and  13  shillings,  so  as 
to  sell  the  compound  at  11  shillings  per  pound:  in  what  propor- 
tions should  they  be  mixed  ?  Ans.  2  at    Ssh, 

to  3  at  \^sh, 

11.  A  trader  bought  one  hogshead  of  rum,  of  115  gallons, 
at  $1.10  per  gallon:  how  many  gallons  of  water  must  be  put 
into  it  to  make  $5,  by  selling  it  at  $1  ? 

Ans.  1Q\  gallons  of  water, 

LESSON  LXIV. 

PERCENTAGE. 

1.  Under  this  head  are  included  interest,  discount, 
commission  and  brokerage,  insurance,  loss  and  gain,  tare 
and  tret,  exchange,  equations  of  payments,  and  in  one 
word,  all  questions  in  which  a  certain  proportional  allow-» 
ance  is  to  be  calculated. 
23 


266  Lesson  lxiv. 

2.  The  rate  of  this  allowance  is  generally  referred  to 
one  hundred,  in  order  to  avoid  small  fractions.  Thus,  for 
certain  services  or  advantages,  an  allowance  of  $5  in 
every  $100  is  made. 

The  7'afe,  in  this  case,  is  said  to  be  5  per  cent.  It  is 
sometimes  referred  to  one,  and  then  is  written  yj^^,  or 
0.05.  The  first  mode  of  expressing  it,  however,  is  more 
general,  though  not  preferable.  We  will  use  both  ways, 
for  practice. 

SIMPLE  INTEREST. 

3.  {  being  the  interest, 
r  the  rate  for  1, 

t  the  time, 
p  the  principal, 
s  the  sum  or  total  amount. 
The  algebraical  formulee    for  interest  are,  i  =  prtj  and  s  =* 

4.  Interest  is  an  allowance  made  for  the  use  of  money 
borrowed,  in  proportion  to  the  time  and  at  a  certain  7'ate 
agreed  upon  between  the  parties,  or  fixed  by  law. 

The  principal  or  capital  is  the  sum  lent,  and  on 
which  the  interest  is  paid. 

The  word  amount  is  particularly  used  to  designate  the 
principal  and  interest  added  together. 

The  interest  of  100  for  one  year  determines  the  rate 
of  interest  ;  it  is  the  rate  per  cent,  per  annum. 

When  the  rate  is  established  by  law,  it  is  styled  the 
legal  rate.  This,  in  most  of  the  United  States,  is  6  per 
cent. ;  and,  therefore,  in  such  states,  when  no  rate  is 
mentioned,  6  per  cent,  is  understood. 

In  New  York,  however,  it  is  7  per  cent.,  and  8  in 
Louisiana. 

Interest  exacted  at  a  rate  exceeding  legal  interest,  is 
considered  usury. 

Interest  may  be  either  simple  or  compound.  We  con^ 
sider  only  simple  interest  in  this  lesson. 

5.  All  questions  in  interest  are  like  the  following  : 

What  is  the  interest  on  $4,500,  for  3  years  and  6  months,  at 
7  per  cent,  per  annum  ? 


SIMPLE    INTEREST.  267 

This  is  evidently  a  question  in  compound  proportion^ 
since  ihe  interest  depends  on  two  ratios  :  1st.  That  of  the 
■principal  to  100  dollars ;  2d.  That  of  the  time  to  the 
unit  of  time,  one  year. 

The  statement  is  made  by  considering,  in  the  first 
place,  that  since  $7  is  given  for  the  use  of  100  dollars 
one  year,  as  many  times  $7  must  be  paid  for  the  use  of 
$4,500  one  year  as  there  are  100  dollars  in  this  capital; 
that  is, 

$7  X  Vo%o ;     or,     $0.07  x  4500. 

And,  in  the  next,  that  this  interest  for  one  year  must 
be  multiplied  by  the  time  during  which  the  money  wa? 
borrowed  ;  making  the  answer 

7  X  Y/o°  X  (V-  ^^0')  =  '7  X  Vo%^  X  3.5  =  $1,102.50 ; 

which  shows  that  interest  is  calciilafed,  in  general,  by 
multiplying  the  capital  by  the  rate  of  interest  and  by 
the  time,  and  dividing  by  100. 

It  must  be  observed  that  the  number  3.5,  which  stands 
for  the  time,  Syr.  dmo,,  or  3J  years,  decimally  expressed, 
is,  in  fact,  the  abstract  number  resulting  from  the  ratio  of 
the  time,  Syr,  6mo.,  to  its  unit  of  comparison,  one  year. 
So  that  the  above  statement  should  in  reality  be 

The  unit  of  comparison  is  generally  omitted  in  the  de- 
nominator, when  its  nature  is  well  understood ;  because 
it  does  not  affect  the  numerical  result,  and  merely 
changes  the  denominate  quantity,  3.5  years,  into  the  ab- 
stract number,  3.5. 

6.  Sometimes  the  rate  of  interest  is  not  reckoned  by 
the  year,  but  by  some  other  unit  of  time  ;  the  month,  for 
example. 

The  difference  in  the  unit  of  time,  however,  produces 
none  in  the  application  of  the  rule.  Thus,  in  the  follow- 
ing question : 

What  is  the  interest  on  $6,000  for  1  year  1  month  and  15  days, 
at  the  rate  of  |  per  cent,  per  month  ? 


268  LESSON    LXIV. 

The  statement  would  be  $|  X  \^«^«  X  ^^^^i^> 

In  which  the  compound  time  may  be  reduced  to  the 
standard  of  comparison,  the  month,  in  order  to  make  the 
ratio  a  simple  abstract  number )  making  thus  the  state- 
ment 

$1  X  ^{-i^  X  13.50  =  §607.50; 

in  which  the  rate,  principal,  and  time,  reduced  to  its 
standard,  are  multiplied  together,  and  divided  b}^  100, 
according  to  the  rule. 

Qiiestio?is. — What  is  percentage  ?  Interest  ?  Rate  of  inte- 
rest ?  Principal  ?  Amount  ?  Legal  rate,  in  general  ?  What  is 
it  in  New  York  ?  In  Louisiana  ?  In  the  other  states  ?  What  is 
called  usury?  What  rule  of  arithmetic  do  questions  in  interest 
belong  to  ?  What  is  the  rule  for  calculating  interest  ?  W^hat 
should  be  done  with  compound  time  ?  How  is  this  factor  consi- 
dered?    Explain  it  by  ratios. 

EXERCISES. 

In  the  following  questions,  it  must  be  recollected  that  begin- 
ners should  always  indicate,  in  a  complete  statement,  all  the 
operations  to  be  performed,  whereby  they  will  very  frequently 
simplify  considerably  their  execution. 

In  most  cases,  it  will  be  best  to  reduce  compound  time  to 
decimals. 

1.  How  much  is  the  interest  of  $187.25,  for  16  months,  at  6  per 
cent,  per  annum  ? 

Statement:  ^^  ^iqq^"^'^  ^   if  =  §  X  1.8725  =  $14.98. 

Here,  you  cut  off  two  figures  of  the  capital,  and  cancel  12, 
with  6,  and  2,  out  of  16. 

2.  What  is  the  interest  of  $694.84,  for  9  months,  at  10  per 
cent,  per  annum?  ^^s-  $52.11  3m. 

or,  $52,113. 

3.  How  much  is  the  amount  (capital  and  interest)  of  $985,  for 
5  years  and  8  months,  at  .06  per  annum  ?  Calculate  the  interest 
for  5§  years,  or  reduce  all  to  months.  Aiis.  $1,319. 90. 

4.  What  is  the  interest  of  $126.46,  for  9  months,  at  6  per  cent.  ? 

5.  Find  the  interest  of  547.80,  for  3  years,  at  0.05  per  annum. 

Ans.  $ 

6.  Find  the  interest  of  $223.20,  for  9  years  and  3  months,  at  7 
per  cent.  ^^*'>-  $144,522. 


SIMPLE    INTEREST.  2 09 

7.  What  is  the  interest  on  $498.50,  for  6  years,  at  4|  per  cent, 
per  annum?  ^?Z5.  $134,595. 

S.  What  is  the  interest  on  $4,988.75,  for  8  years,  at  0.08  per 
annum?  A?is.  $3,192.80. 

9.  What  is  the  interest  of  $912,  for  Syr.  5mo.  ISda.,  at  5  per 
cent?  A?is,  $158.08. 

10.  What  is  the  interest  on  $648,  for  lyr.  lOmo.  19da.,  at  0.10  ? 

A7is.   $122.22. 

11.  A  merchant  has  purchased  cloth  for  $3,859.25;  for  which 
he  offers  a  note,  payable  in  18  months,  including  interest  at  |  per 
cent,  per  month :  what  should  be  the  amount  on  the  face  of  the 
note  ?  Ans.  $4,380.25. 


LESSON  LXV. 

1.  Although,  with  a  little  practice,  simplifications  ap- 
plicable to  particular  cases  will  be  readily  discovered, 
there  are  some  methods,  which  recur  so  frequently  in 
practice,  that  they  may  with  propriety  be  introduced 
here. 

Simple  interest,  for  a  number  of  months,  is  readily 
calculated,  when  the  rate  per  cent,  is  an  aliquot  part  or 
factor  of  12  J  that  is,  when  it  is  6,  4,  or  3  per  cent. 

INTEREST    AT    SIX   PER    CENT.    PER   ANNUM. 

2.  Let  it  be  proposed  to  find  the  interest  of  $375,  for 
2  months,  at  6  per  cent.     The  statement  would  be 

$375  X  0.06  X  2  $375X6X2 

12  '  ^^       100X12      — 'P^-'^- 

In  which  6x2  cancels  12;  and  the  interest  for  two 
months  is  thus  obtained  by  simply  cutting  off  two  figures 
of  the  capital ;  and,  from  this,  the  interest  for  any  nuM' 
her  of  months  may  be  calculated^  by  multiplying  that  for 
two  months  by  half  the  number  of  months. 

Thus  would  the  interest  of  the  above  sum,  $375,  foi 
8  months,  be  .         .         $3.75  X  4    =  $15.00. 

For  9  months,  .         $3.75  X  4^  =  $16,875. 

23* 


270  LESSON   LXV. 

INTEREST    AT    FOUR    PER    CENT.    PER    ANNUM. 

3.  If  it  were  asked  to  get  the  interest  of  §4^48.50,  for 
3  months,  the  statement  would  be 

$448.50X4X3  _ 

100  X  12         -  ^^-^^S^. 

Which  shows  that,  at  4  j)er  cent,  per  annum,  the  interest 
for  three  months  is  obtained  by  merely  removing  the 
unites  point  two  places  to  the  left. 

For  any  other  time^  the  interest  would  be  found  by 
multiplying  by  one  third  of  the  number  of  months^  and 
displacing  the  decimal  point  two  places  to  the  left. 

INTEREST    AT    THREE    PER    CENT.    PER    ANNUM. 

4.  The  interest  for  four  months^  at  3  per  cent,^  icould 
he  found  by  displacing  the  decimal  point  tico  places  to 
the  left. 

For  any  number  of  months,  the  interest  would  be  ob- 
tained, after  having  displaced  the  units'^  point,  by  multi- 
plying by  one  fourth  the  number  of  months. 

Example:  What  is  the  interest  on  $4,768,  for  16 
months,  at  3  per  cent,  ?        Ans,  $47.68  x  4  ==  $190.72. 


What  is  it  for  9  months  ? 


Ans.  $47.6^8X9  ^$107.28. 


INTEREST    FOR    A    NUMBER    OF    DAYS. 

5.  In  the  computation  of  interest  in  Federal  money,  it 
is  customary  to  reckon  the  year  at  12  months,  and  each 
month  at  30  days  5  considering  thus  the  year  to  contain 
only  360  days  instead  of  365  days.  This  method  makes 
but  an  inconsiderable  difference  in  the  amount  of  interest, 
and  is  very  convenient  in  practice,  especially  at  rates 
like  6,  4,  3,  9,  which  are  factors  of  360.  Thus,  for  an 
amount  of  $696,  at  6  per  cent,  per  annum,  for  37  days, 
we  would  2:et 

4.292. 


X37X6  _  0.696  X  37 

360X100      ~-        6 


SIMPLE    INTEREST.  271 

Which  shows  that,  to  get  the  interest  for  a  niimher  of 
days,  at  6  per  cent,  per  annum,  yon  must  cut  off  three 
figures^  multiply  by  the  mumher  of  days,  and  divide  by  6. 

At  4  per  cent.,  you  would  have  to  divide  by  9. 

At  3,  by  12. 

At  9,  by  4. 
And  at  12,  by  3. 

6.  The  most  usual  rate  of  interest  being  6  per  cent., 
other  rates  are  frequently  calculated  from  it,  on  account 
of  the  facility  with  which  it  is  computed. 

5  per  cent.,  by  subtracting  -j-  from  it ;  7,  8,  and  9,  by 
adding  respectively  ],  -i,  >,  to  the  interest  at  6  per  cent. 

Example:  $156,  at  7  per  cent.,  for 4-  months,  =  3.12 
4-  3:^U)  =  3.64. 

7.  Simplifications  of  this  kind,  with  a  little  practice, 
will  be  readily  discovered  in  all  cases,  without  reference 
to  stated  rules. 

8.  In  general,  questions  which  involve  years,  months, 
and  days,  are  performed  as  in  compound  numbers,  which 
the  preceding  remarks  serve  to  simplify,  in  the  cases  re- 
ferred to. 

EXAMPLES. 

I.  What  is  the  interest  of  $418,  for  1  year  7  months  and  17 
days,  at  6  per  cent.  ? 


$418 

Ij/r,  Imo.   llda. 

For  1  year, 

25.08 

For  6  months, 

12.54 

For  1  month, 

2.09 

For  15  days,  |  of  1  month, 

1.045 

For  2  days,    i    of  1  month, 

0.139  + 

$41,894-}- 

II.  How  much  is  the  interest  of  $268.44,  for  3  years  5  months 
26  days,  at  6  per  cent,  per  annum  ? 

For  3  years,  18  times  2.6844,  .         .  $48.3192 

For  5  months,  |  or  V°  of  do.  .  6.7110 

For  15  days,  -j'q  of  5  months,  •  .6711 

For  10  days,  y^  of  5  months,  .  .4474 

For  1  day,  -^q  of  preceding,  .         .  .04474 

$56.19344 


272 


LESSON    LXV. 


III.  What  is  the  interest  of  $910.50,  for  3  years  9  months  26 
days,  at  7  per  cent.  ? 

For  3  years,      .         .         21  X  9.1050  = 
For  9  months,  ^  of  3  years, 
For  18  days,  yj  of  9  months, 

For  6,     I,  

For  2,     i,  


Or  else, 

For  1  year,     . 
For  3  years, 
For  6  months, 
For  3       " 
For  15  days. 
For  10  days, 
For  1  day, 


$191.2050 
47.80125 
3.1S675 
1.06225 
.35408i 

$243.60933^ 


$9.1050X7  =  63.7350. 


^  of  1  year, 


•§-  of  3  months, 


$191.2050 
31.8075 
15.93375 
2.655625 
1.770416'6 
.177041%' 

$243. 60933*37 


Questions. — How  is  the  interest  on  Federal  money,  at  6  per 
cent.,  calculated  for  2  months  ?  For  any  number  of  months  ?  At 
4  per  cent,  for  3  months  ?  For  any  number  ?  At  3  per  cent,  for 
4  months  ?  For  any  number  of  months  ?  For  a  number  of  days, 
how  is  it  customary  to  reckon  the  year  ?  How  is  the  interest 
for  a  number  of  days  calculated,  at  3  ?  4  ?  6  ?  9  ?  12  per  cent.  ? 
How  may  interest  be  calculated  for  any  rate,  from  that  at  6  per 
cent.  ?     How,  when  the  time  is  a  compound  number.? 

EXERCISES. 

1.  What  is  the  interest  of  $230.50,  for  9  months,  at  0.04  per 
annum?  Ans, 

2.  On  $95,  for  2  months,  at  6  per  cent.  ?  Ans» 

3.  On  $194,  for  4  months  and  12  days,  at  0.06? 

Ans,  $4,268. 

4.  On  $265.48,  for  2  months  and  21  days,  at  6  per  cent.  ? 

Ans,  $3,584. 

5.  On  $318,  for  10  months  and  16  days,  at  0.06  per  cent.  ? 

Ans.  $16,748. 

6.  On  $1,  for  18  days,  at  6  per  cent.  ? 

7.  On  $47,984,  for  15  months,  at  3, 4,  6  per  cent,  per  annum? 

8.  On  $618.96,  for  41   days,  at  3,  4,  6,  9,  12  per  cent,  per 
annum  ? 

9.  Oa  $240,  for  2  years  6  months  and  13  days,  at  0.06? 

Ans,  $36.52. 


SIMPLE    INTEREST.  273 

10.  On  $615.75,  for  2  years  7  months  and  16  days,  at  4;  5;  6; 
7;  8;  9;  10  per  cent,  per  annum  ? 

11.  On  $241.60,  for  3  years  4  months  and  15  days,  at  6  pel 
cent..?  A?is.  $48,924. 

12.  On  1,440,  for  5  years  11  months  and  28  days,  at  0.08  ? 

A71S.  $690.56. 


LESSON  LXVI. 

1.  In  the  preceding  lesson,  we  have  considered  the 
year,  according  to  a  general  practice,  to  consist  of  360 
days.  Some  persons,  however,  calculate  by  days  at  the 
more  correct  rate  of  365  in  a  year. 

Example. — What  is  the  interest  on  $7,086,  at  6  per  cent.,  for 
39  days  ? 

Statement:  .         $7,086  X  yfo  ^  At  =  $45.43. 

It  will  be  readily  perceived  that  this  mode  of  com- 
puting, which  is  less  favorable  to  the  lender,  is  not  so 
simple  in  practice,  and  admits  only  of  occasional  cancel- 
lings. 

The  interest,  in  this  case,  will  be  63  cents  less  than 
when  the  year  is  considered  as  only  360  days. 

2.  When  the  sum  on  which  the  interest  is  to  be  calcu- 
lated, is  in  pounds,  shillings,  and  pence,  the  operation  is 
not  so  simple  as  in  Federal  money. 

3.  When  the  fractions  are  convenient,  you  may, 

I.  Reduce  the  shillings  and  pence  to  decimals  of  a 
pound. 

II.  Then  find  the  interest^  as  above. 

III.  Andy  finally,  reduce  the  decimal  part  of  the  an- 
swers to  shillings  and  pence. 

Example  \st. — What  is  the  interest  on  X150  15*^.  6J.,  at  6 
per  cent.,  for  2  years  and  6  months  ? 

The  principal,  in  decimals,  is  £150.775,  and  the  interest  on  it 
jei.50775  X  15  ==  £22.61625  =  £22  \2sh,  3'id. 

4.  In  other  cases,  when  both  the  time  and  principal 

S 


274 


LESSON   LXVI. 


being  compound  numbers,  the  fractions  are  not  readily 
changed  into  decimals,  it  will  be  found  shorter  to  per- 
form the  operation  by  aliquot  parts.  The  first  step  of 
the  operation  will,  in  general,  be  to  multiply  the  rate  per 
cent,  by  either  the  capital  or  time,  as  may  appear  more 
simple. 

Example  2d. — What  is  the  interest  on  <£70  19sA.  4c?., 
at  6  per  cent.,  for  2  years  8  months  and  16  days? 

Statement :  .         (£70  \Qsh.  Ad.)  X  G  X  {2yr.  Svw.  16da.) 

]    Too 

In  this  case,  it  will  be  best  to  multiply  the  capital  by 
6,  and  then  the  operation  is  reduced  to  multiplying 

£425    IQsh. 

by  y^  of         •         •         •         •  Syr.  Smo,  16da, 

\ 
For  2  years,  take  -g^jj 
8  months,  § 

16  days,  j\ 

£11    lOsk.  10.5381 

5.  If  the  year  w^as  reckoned  at  365  days,  the  interest 
for  the  16  days  should  be  taken  separately,  as  follows : 

(£425  16sk.)  X  16        ^425.80  X  16  _  68.1280  _  .  ,    ^  ^..i  , 
100X365 ==      100X365      "  -"365-  "  ^'^'  ^''^^--^' 

Showing  a  difference  of  only  0.622  of  a  penny  between 
the  two  modes  of  reckoning  interest  for  16  daysj  a  dif- 
ference too  inconsiderable  to  be  noticed. 

Example  3d. — To  find  the  interest,  at  7^  per  cent,  per 
annum,  of  £119  7sh.  Id.,  for  4  years  7  months  and  23 
days. 

Here,  instead  of  multiplying  either  the  time  or  capital 
by  the  rate,  it  will  be  found  easier  to  calculate  first  the 
interest  for  one  year. 

And  then  to  get  from  it,  by  aliquot  parts,  that  for  the 
given  time. 


£8    10^-^. 

3.84J. 

2    16 

9.28 

0      3 

9.418f 

SIMPLE   INTEREST.  275- 

OPERATION. 

;eil9     Ish.     Id. 
^yr.  Ittw.  23da» 

£5  19sh.  4.55d, 


5(J 
For  21,         "  i     .        2    19       8.275 


Interest  for  1  year,  .     £S  19sh. 

"  for  4  years, 

"  for  6  months,  |, 

"  for  1  month,  i  ^ 

«  for  15  days,  |, 

"  for  5     "  - 

«  for  3     "  y^  of  1  month, 


0.825«f 

£35 

16^//. 

3.3d. 

4 

9 

6.4 

0 

14 

11.1 

0 

7 

5.5 

0 

2 

5.8 

0 

1 

5.9 

:e41 

12sh 

2.0d. 

Questions. — How  should  interest  be  calculated,  if  the  year  be 
reckoned  at  365  days  ?  To  whom  is  this  mode  of  computing 
favorable  ?  How  is  interest  on  sterling  money  calculated,  when 
the  fractions  are  simple  ?     In  other  cases  ? 

EXERCISES. 
The  year  being  considered  365  days : 

1.  What  is  the  interest  on  $87.56,  for  72  days,  at  6  per  cent, 
for  a  year  ?  A^is.  $1,036. 

2.  On  $2,962.19,  for  254  days,  at  6  per  cent.      Ans.  $123,678. 

3.  On  $1,733.97,  for  102  days,  at  8  per  cent.  ?        Aiis. 

4.  On  .£355  155/^.,  for  4  years,  at  4  per  cent.  ? 

Ans.  £5Q  \Ssh.  4|^. 

5.  On  £32  5sh.  8d.,  for  7  years,  at  4  J  per  cent.  ? 

Ajis.  XIO  3sh.  Ad. 

6.  On  je319  Qd.,  for  5f  years,  at  3f  per  cent.  ? 

A?is.  £68  I5sh.  9|^. 

7.  The  year  being  365  days,  what  is  the  interest  on  X'107,  for 
117  days,  at  4|  per  cent,  per  annum?  A?is.  £1  \'2sh.  Id, 

8.  On  £\1  5slu,  for  117  days,  at  4f  per  cent.  ?     Ans,  5sh,  3d. 

9.  On  £712  65^.,  for  8  months,  at  7 A  per  cent.  ? 

Ans,  £35  12sk,  3|^. 

10.  The  year  being  360  days,  on  £107  165^.  10^.,  at  5  per 
cent.,  for  7  years  and  12  days?  Ans.  £37  I8sh.  5d. 

11.  How  much  is  the  amount  of  $298.59,  from  the  19th  May, 
1817,  to  the  nth  August,  1848,  at  5|  per  cent.  ?      Aits.  $319.67. 


276  LESSON   LXVTI. 

12.  Uichmond^  June  14  th,  1846. 

For  value  received,  I  promise  to  pay,  on  the  29th  day  of  April, 
1847,  to  Henry  Lee  or  order,  the  sum  of  one  hundred  and  ninety- 
six  dollars,  bearing  interest  from  date,  at  5f  per  cent. 
$196.  James  Ogden. 

What  is  the  amount  on  day  of  payment  ?  Ans.  $205.86. 

13.  RichmoJid,  January  9th,  1845. 
$658. 

Nine  months  after  date,  I  promise  to  pay  to  Thomas  Wat- 
kins  or  order,  with  interest  from  date,  at  6  per  cent,  per  annum, 
six  hundred  and  fifty-eight  dollars,  for  value  received. 

Jeremiah  Nichols. 
Amount, $ 


LESSON  LXVIL 

1.  Having  now  explained  different  methods  of  com- 
puting interest,  it  may  not  be  amiss  to  complete  the  sub- 
ject with  several  questions  which  simple  interest  may 
give  rise  to. 

2.  It  will  be  readily  perceived  that  the  'principal^  the 
rate,  the  interest,  the  time,  and  the  amount,  are  so  con- 
nected that  any  one  of  them  may  be  obtained  by  means 
of  the  others. 

CASES   I.   AND   II. 

3.  The  principal,  rate,  and  time  being  known,  to  find 
the  interest  and  the  amount. 

These  cases  form  the  subject  of  the  preceding  lessons. 

CASE   III. 

4.  The  principal,  rate,  and  interest  being  given,  to  find 
the  time.     Algebraical  formula  (see  LXIV.) '-  t  =  ^^ 

Divide  the  given  interest  by  the  interest  for  one  year, 
since  the  interest  is  the  product  of  the  interest  for  one 
year,  by  the  time. 

Example. — How  long  must  the  interest,  at  8  per  cent;* 


SIMPLE    INTEREST.  277 

on  §560  be  collected  by  a  certain  individual,  in  order 
that  he  may  be  paid  a  debt  of  §106.40  ? 

o.  .  *        $106.40 

560  X  0.08        '^^^'  ^^^°'  ■^'^'^^* 

The  numerator  is  the  aggregate  interest  to  be  collected ; 
the  denominator,  the  interest  of  §560  for  one  year,  at 
the  rate  0.08,  or  8  per  cent. 

CASE    IV. 

5.  The  principal,  interest,  and  time  being  given,  to 
find  the  rate.     Formula:  r  ==  ~ 

Divide  the  interest  by  the  time  and  by  the  principal ; 
since  the  aggregate  interest  is  the  product  of  the  princi- 
pal and  time  by  the  rate. 

Example. — A  person  has  borrowed  §3,750,  for  2  years 
and  6  months,  and  paid  for  the  use  of  it  §719.25 :  at 
what  rate  did  he  pay  interest  ? 

^719  25 
Statement:  3^750  ^  2^7,  =  0.0767,  or  7.67  per  cent.; 

an  illegal  interest  in  many  states. 


CASE   V. 

6.  The  time,  rate,  and  interest  being  given,  to  find  the 
principal.     Formula :  p  =  ■^. 

Divide  the  interest  by  the  time  and  the  rate;  since  the 
interest  would  be  produced  by  multiplying  together  the 
capital,  rate,  and  time. 

Example, — On  what  principal  should  the  interest  be 
given  up  for  16  months,  so  as  to  pay  up  a  debt  of  §124, 
the  rate  of  interest  being  0.06  ? 

statement :  --^^  =  IHf^  =  $1,550. 
0.06  X  if  8 

24 


278 


LESSON    LXVI. 


N.  B. — If,  in  this  case,  the  amount  instead  of  the  principaK 
were  required,  find  the  principal  first  and  add  it  to  the  interest. 

CASE  VI. 

7.  In  some  questions,  the  amount  may  be  given :  if, 
then,  either  the  principal  or  the  interest  is  also  known, 
the  other  is  readily  obtained  by  subtracting  the  known 
one  from  the  amount ;  and  thus  the  question  will  fall 
under  one  of  the  preceding  cases :  as,  for  example,  in  this 

Question. — A  man,  for  the  loan  of  $200,  requires  a 
note  of  $500,  payable  in  10  years:  what  is  the  rate  of 
interest  thus  virtually  exacted  ? 

The  difference,  $300,  is  the  interest;  and,  by  Case  IV., 

300        __(..- 
10X200  —  ^•■^^' 

or  15  per  cent,  the  rate  required. 

8.  The  only  case  which  requires  a  separate  investiga- 
tion, is  that  when 

The  time,  rate,  and  amount  being  given,  it  is  required 
(o  find  the  interest  and  the  prijicipaL 

This  case  is  properly  the  rule  of  discount,  which,  on 
account  of  its  importance,  we  will  consider  in  a  separate 
lesson. 

Questions. — The  principal,  rate,  and  time  being  given,  how  is 
the  interest  found?  How,  the  amount?  The  principal,  rate,  and 
interest  being  given,  how  is  the  time  found?  The  principal,  in- 
terest, and  time  being  given,  how  is  the  rate  found  ?  Having 
given  the  time,  rate,  and  interest,  how  is  the  principal  found  ? 
How,  the  amount  ?  If  the  interest  and  capital  are  known,  how 
is  the  amount  found?  How  would  you  proceed,  in  questions  in 
which  the  amount  and  either  the  principal  or  the  interest  ar^ 
given  together,  to  find  the  other  ?  What  rule  do  questions  belong 
to,  in  which  the  time,  rate,  and  amount  are  given,  and  either  the 
interest  or  principal  required  ? 

EXERCISES. 

1.  A  person  paid  $94,902  on  $981.75,  at  6  per  cent,  per  annum : 
what  was  the  time  ?  ^         Ans.  \yr.  Ivio.  lOda. 

2.  In  what  time  v/ill  $1,800  amount  to  $1,853.70,  at  6  per 
cent.  ?  Ans,  5mo.  '29da» 


SIMPLE    DISCOUNT.  279 

8.  ^423.20  have  been  paid  for  the  loan  of  $920,  for  8  years 
what  is  the  rate  ?  Ans.  0.0575,  or  5|  jper  cent. 

4.  For  a  sum,  $3,650,  a  man  exacts  a  note  of  $4,790. 62|,  pay- 
able in  2  5  years ;  what  is  the  rate  of  his  usury  ? 

Ana,    12 1  per  cent, 

5.  £9  125^.  \d.  have  been  paid  for  the  use  of  £32  ^sh.  8cL, 
for  7  years :  what  is  the  rate  of  interest  ? 

6.  For  the  loan  of  £873  15-5^.,  for  2J  years,  a  note  of  £977 
lOsh.  l^d.  is  taken:  is  the  rate  usurious?  Ajis.  A^  per  cent, 

7.  $922.14  having  been  paid,  at  an  interest  of  8  per  cent.,  for 
3  years  and  11  months,  what  was  the  capital  ?  Ans.  $2,943. 

8.  The  interest  of  a  note  bearing  interest  at  5  per  cent., 
amounts,  in  3  years,  to  £82  osh.  3d, :  what  was  the  sum  on  the 
face  of  the  note.^  Ans.  £547  I5sh. 


LESSON  LXVIII. 

SIMPLE  DISCOUNT. 

Algebraical  formulae  of  discount:  p  = ,  d  =  s  — ^,  or 

srt  l-\-rt 

'^  =  i+TE  (S««  LXIV.,  3) 

1.  Discount  is  a  deduction  made  from  a  debt  when  it 
is  paid  before  it  becomes  due. 

Discount  may  be  like  interest,  simple  or  compound.  In 
this  lesson,  we  consider  only  Simple  Discount.  Compound 
Discount  will  be  the  subject  of  a  subsequent  lesson. 

If  I  owe,  for  instance,  a  note  of  $106,  payable  at  the 
end  of  the  year,  and  I  am  asked  to  discharge  this  debt 
now,  it  cannot  be  expected  that  I  would  agree  to  it  with- 
out receiving  some  advantage  for  this  anticipation  of  pay- 
ment ;  and  this  advantage  should  be  exactly  equal  to  the 
regular  interest  I  would  receive  by  retaining  the  use  of 
the  money  I  pay  out.  It  is  in  fact  the  same  thing  as  if 
I,  in  my  turn,  were  to  lend  money  to  my  lender  to  can- 
cel his  claim  of  $106.  In  this  case,  the  principal  I  give 
him  must  be  such  that,  together  with  the  interest  for  the 
time  the  note  has  to  run,  the  aggregate  amount  may  be 
precisely  equal  to  the  sum  due  at  the  expiration  of  the 
time. 

If  the  interest  is  at  6  per  cent.,  the  note  being  for 
$106,  the  prmcipal  would  evidently  be  100. 


280 


LESSON   LXVIII. 


2.  Hence,  the  rule  of  discount  consists  in  finding  what 
principal  ivould  produce  the  amount  due  in  a  given  time. 
This  is  the  present  value  of  the  note. 

The  discount  is  the- difference  heticeen  the  face  of  the 
note  and  its  present  value. 

This  rule,  therefore,  as  was  said  before,  is  the  same  as 
Case  VI.,  of  the  preceding  lesson ;  where, 

The  amount,  rate,  and  time  being  given,  it  is  proposed 
to  find  the  principal. 

Divide  the  given  amount  by  the  amount  of  one  dollar, 
at  the  given  rate,  and  for  the  given  time. 

Example, — What  is  the  present  worth  and  discount  of 
a  note  of  $450,  due  6  months  hence,  at  6  per  cent.  ? 

In  the  first  place,  it  is  evident  that  $1  would  become 
$1.03  in  6  months,  and  consequently  as  many  dollars 
must  be  paid  now  as  the  num.ber  of  times  the  amount, 
$450,  contains  1.03  j  that  is, 

-4^^  =  $436,893,  and,  for  the  discount,  450  —  436.89 
=  $13,107. 

3.  The  same  may  be  obtained  by  proportions,  as  fol- 
lows: 

Since  $100,  with  the  addition  of  interest,  amounts  to 
$103,  in  6  months,  we  have  the  proportion 

103  :  100  :  :  450  :  fS^X  103^ 
100 

The  correctness  of  the  operation  may  now  be  verified 
by  calculating  the  interest  on  $436,893  and  the  amount, 
which  would  be  found  to  be  $450. 

Questions. — What  is  discount  ?  What  case  in  questions  on  in- 
terest does  it  belong  to  ?  How  is  the  capital  found  ?  How,  the 
discount  ?     By  division  ?     By  proportions  ? 

EXERCISES. 

1.  What  is  the  discount  of  jCSOS  I5sh.,  due  in  18  months,  at 
8  per  cent,  per  annum?  Arts,  X33  Ish.  ltd. 


BANK   DISCOUNT.  SSl 

2.  "What  is  the  present  worth  of  $5,150,  due  in  4|  months,  at 
8  per  cent,  per  annum,  and  allowing  1  per  cent.,  besides,  for 
prompt  payment?  A>is.  $4,950. 

3.  A  is  to  pay  $5,927  on  the  19th  of  April,  and  $5,989  on  the 
19th  of  July  following.  He  prefers  to  pay  all  on  the  19th  of 
January  :  how  much  will  discharge  both  sums,  at  8  per  cent.  ? 

A?is.  RecloJiing  by  months,        ,  $11,569,438. 

By  days,  and  365  in  the  year,  $11,572,820. 

4.  What  is  the  discount  on  $4,590.39,  at  6  per  cent.,  for  47  days 
of  a  year  of  365  days  ?  Ans,  $35.19. 

5.  What  is  the  principal  of  the  amount  $2,202.34,  for  125 
days,  the  year  of  365  days,  at  6  per  cent.  ?  Ans.  $2,158. 

6.  What  is  the  present  value  of  a  note  for  $4,850,  payable  in 
13|  months,  at  |  per  cent,  per  month '/  Ans.  $4,404.09. 

7.  What  is  the  discount  on  a  note  for  $2,850.45,  payable  in 
2  years  and  8  months,  at  8f  per  cent,  per  annum  ?    Ans.  $539.27. 

8.  Bought  goods  to  the  amount  of  $950,  at  90  days :  what 
ready  money  will  discharge  it,  the  interest  being  at  6  per  cent.  ? 

A71S.  ^935.15,  for  3  mo7iths. 
$935.95, /or  90  days. 

9.  What  is  the  present  value  of  three  not'es,  payable  at  one, 
two,  and  three  years,  interest  at  5  per  cent.,  each  note  being  for 
$478.17?  Ans.  $1,305.90. 

10.  What  amount  of  ready  cash  will  discharge  a  debt  of  $1,950, 
of  which  $190  are  payable  in  6  months;  $270  in  one  year;  $490 
in  18  months,  and  $1000  in  2  years,  interest  at  6  per  cent.  ? 

Ans.  $1,781,581. 


LESSON  LXIX. 
BANK  DISCOUNT. 
Formulge  :  p=  s{l  —  rt)  .         .        d  =  srt. 

1.  The  above  method  of  calculating  discount  is  the 
only  one  by  which  neither  party  suffers  a  loss.  It  is, 
however,  not  always  the  discount  allowed  in  ordinary 
transactions,  and  especially  by  bankers. 

2.  When  money  is  obtained  from  a  bank,  it  is  cus- 
tomary to  deduct  and  retain  as  discount  the  interest  on 
the  amount  of  the  note  for  the  time  it  has  to  run,  and 
also  for  thj'ee  days  of  grace  y  which  are  usually  allowed. 

24* 


282 


LESSON   LXIX. 


Thus,  if  a  note  for  $100  be  discounted  at  a  bank,  for 
30  days,  the  interest  for  33  days,  which,  at  6  per  cent., 
is  55  cents,  is  deducted  for  discount :  and  the  holder  of 
the  note  receives  $99.45  for  the  $100  which  appears  on 
the  face  of  the  note. 

3.  This  manner  of  discounting  is  evidently  to  the  ad- 
vantage of  the  banker ;  since  he  exacts  interest  for  more 
than  the  sum  he  advances;  that  is,  not  only  for  the  mo- 
ney you  receive,  but  likewise  on  the  interest  he  with- 
holds.    So  that  you  give  up  the  interest  of  the  interest. 

To  make  this  perfectly  clear  by  a  simple  case,  let  us 
suppose  that  you  hold  a  note  for  $106,  payable  in  one 
3^ear  :  it  is  plain  that  it  is  equivalent  to  $100  now;  yet 
the  banker  deducts  the  interest  not  only  on  the  $100,  but 
also  on  the  $6,  which  are  the  interest  to  accrue  during 
the  year. 

This  shows  that  the  discount^  and  not  the  interest^  is 
the  just  deduction  for  anticipation  of  payment.  Yet  the 
last  method  is  in  common  use,  especially  for  small  ac- 
counts, not  only  because  it  is  more  expeditious,  but  pro- 
bably also  because  the  lender  has  the  power  to  make  his 
own  terms. 

This  mode  of  discounting  being  only,  after  all,  a  mere 
calculation  of  simple  interest,  it  is  unnecessary  to  give 
any  example  or  exercise  on  it. 

4".  As  to  discount  allowed  for  prompt  payment,  or 
other  considerations  which  have  no  reference  to  time,  the 
following  simple  examples  will  suffice : 

1.  How  much  is  the  discount  of  $853,  at  2  per  cent.  ? 

Ans.  $17.06. 

2.  An  agent  sells  property  for  $985.75,  and  is  to  receive  4  per 
cent,  for  his  tiouble  :  how  much  is  he  to  pay?         A7is.  $946.32. 

COMMISSION  OR  BROKERAGE,  AND  INSURANCE. 

5.  Commission  or  Brokerage  is  an  allowance  made 
to  an  agent  for  buying  or  selling.  The  allowance  is  ge- 
nerally a  certain  percentage  or  rate  per  hundred,  on  the 
amount  paid  or  received. 

6.  Insurance  is   an  allowance  of  a  certain  rate  per 


COMMISSION    AND    INSURANCE.  283 

cent.,  made  to  an  individual  or  company,  who,  in  consi- 
deration tfiereof,  will  make  good  the  loss  sustained  by 
fire,  navigation,  storms,  &c.,  up  to  the  amount  or  risk 
insured  for. 

The  written  agreement  is  called  the  policy. 
The  sum  paid  for  insurance  is  the  'premium. 
The  allowance  made,  in  all  these  cases,  is  readily  cal- 
culated by  the  preceding  rules  of  interest. 

Questions. — What  is  the  usual  rule  of  discount  of  bankers  ? 
In'Avhat  does  it  differ  from  true  discount?  What  is  commission? 
Brokerage  ?     Insurance  ?     Policy  ?     Premium  ? 

EXERCISESi 

i.  What  is  the  commission  on  $2,716.50,  at  2|  per  cent.  ? 

Ans.   $54,412. 

2.  The  sale  of  certain  goods  amounts  to  $1,873.40  :  what  is  to 
be  received  for  them,  allowing  2|  per  cent,  for  commission,  and 
{  per  cent,  for  prompt  payment  of  the  net  proceeds  ? 

^?i5.  $1,821,599. 

3.  What  is  the  present  worth  of  $9,150,  due  in  7|  months, 
discounting  at  the  rate  of  5  per  cent,  per  annum,  and  allowing 
]|  per  cent,  for  prompt  payment?  Ans, 

4.  What  is  the  commission  on  £l,371  9^^.  3^.,  at  5  per  cent.? 

A71S,  £68  llsh.  5\di 

5.  What  is  the  brokerage  on  $1,853,  at  f  per  cent.  ? 

A71S.  $13,897. 

8.  «  "  on  £874  \5sh.  3cl.,  at  J  per  cent.  ? 

A71S.  £2  3sk.  8ld, 

7.  «  «  on  £1,321  llsh,  Ad.,  at  1|  per  cent.  ? 

A71S.  £14  llsh.  4d, 

8.  A  factor  receives  $988,  to  lay  out  in  goods,  allowing  his 
commission  of  4  per  cent,  on  the  purchase :  how  much  does  he 
lay  out  ?  Ans.  $950. 

9.  A  factor  has  in  his  hands  $3,960,  which  he  is  desired  to  lay 
out  in  iron,  reserving  2h  per  cent,  on  the  purchase ;  the  iron  being 
at  $95  per  ton,  how  much  can  he  purchase  ? 

A?i6.  31 T.  llcwt.  3qr.  16^-^3. 

10.  What  is  the  premium  on  $1,873,  at  |  per  cent.  ? 

11.  "  "         on  £924,  at  7  per  cent.  ? 

A71S.  £64  13^7^.  Id. 

12.  What  must  be  paid  for  insurance  on  a  house  valued  at 


284 


LESSON   LXX. 


$7,000,  the  cost  of  the  policy  being  SI. 50,  and  the  rate  of  in' 
suring  i  per  cent,  j  and  the  insurance  being  effected  on  only  |  of 
the  valuation  ?  Ans, 


LESSON  LXX. 
TARE  AND  TRET. 

1.  These  are  allowances  made  in  selling  goods  by 
weight. 

Tret  is  an  allowance  to  the  buyer  for  waste,  dust,  &c., 
generally  at  the  rate  of  4  pounds  per  104  pounds. 

2.  Tare  is  a  deduction  for  the  weight  of  the  case  or 
envelope  containing  the  commodity. 

3.  Draft  is  an  allowance  on  the  gross  weight  j  it  is 
always  deducted  before  the  tare. 

4.  Gross  iceigJit  is  the  whole  weight  of  the  commo- 
dity, together  with  the  hogshead,  barrel,  bag,  or  box, 
which  contains  it. 

5.  Net  weight  is  what  remains  after  the  allowances 
have  been  deducted,  and  which  is,  paid  for. 

EXAMPLES. 

1.  At  $1.25  per  pound,  what  will  3  chests,  of  hyson  tea  come 
to,  weighing  96/3.,  97/5.,  and  101/5.;  the  tare  being  20/5.  per 
chest  ? 

96  +  97  -f-  101  _  3  X  20  =  234/5.,  and  Sl.25  X  234  =  $292.50. 

2.  What  is  the  cost  of  a  firkin  of  butter  weighing  gross  132/5., 
at  19  cents ;  the  tare  being  15/5.  for  the  first  50/5.,  and  1/5.  for 
every  ten  pounds  over  fifty  ?  Ans.  $20.71. 

N.  B. — Fractions  less  than  5  pounds  are  not  considered ;  but, 
from  5  up  to  10,  1  is  added  to  the  tare :  so  that,  if  the  firkia 
were  129  instead  of  132  pounds,  the  tare  would  be  the  same. 

PROFIT  AND  LOSS. 

6.  This  is  an  application  of  the  rule  for  interest,  by 
which  the  amount  gained  or  lost  in  a  mercantile  trans- 
action is  ascertained,  and  which  also  serves  to  determine 


PROFIT   AND   LOSS.  285 

at  what  price  goods  should  be  sold  to  realize  a  certain 
profit  or  sustain  a  certain  loss. 

These  operations  require  no  new  rule,  and  a  few  ex- 
amples will  suffice. 

Questions. — What  is  tare;  tret;  draft;  gross  weight;  net 
weight  ?  What  is  profit  and  loss  ?  How  is  it  calculated  ?  How 
do  you  find  the  rate  per  cent,  of  either  profit  or  loss  ?  How 
would  you  find  at  what  price  an  article  must  he  sold,  to  lose  or 
gain  a  certain  percentage  on  the  first  cost  ?  The  price  an  article 
sold  for,  and  the  percentage  realized  or  lost  being  known,  how 
would  you  find  the  first  cost  ? 

EXERCISES. 

1.  I  purchased  80  yards  of  cloth,  at  $3.75  per  yard,  and  sold 
at  $4.50 :  how  much  did  I  gain,  and  what  is  it  per  cent,  on  the 
purchase  ? 

2.  A  trader  bought  one  hogshead  of  strong  rum,  containing  115 
gallons,  at  $1.10  per  gallon :  how  many  gallons  of  water  must  be 
put  in  it  to  gain  $5,  by  selling  it  at  $1  per  gallon  ? 

3.  A  merchant  bought  2,750  bushels  of  wheat,  for  S3, 300; 
but,  finding  it  damaged,  he  agreed  to  sell  it  at  a  discount  of  10 
per  cent. :  what  will  it  be  per  bushel?  Ans.  ^1.08. 

4.  A  merchant  sells  his  goods  at  2?  pnce  profit  on  the  shil- 
ling :  how  much  is  it  per  cent.  ? 

5.  Bought  a  piece  of  baize,  of  42  yard?,  for  ^4  lish.  6o?,,  and 
sold  it  at  2sh.  6d.  per  yard  :  what  is  the  profit  or  loss  ? 

Alls.   lOsk.  6d.  projlt. 

6.  Which  is  the  better  bargain,  in  purchasing  fish  at  17  shillings 
per  quintal,  at  4  months  credit,  or  I6sk.  Sd.,  cash;  interest  at 
6  per  cent,  per  annum? 

7.  Bought  4  hogsheads  of  rum,  containing  450  gallons,  at  $1 
per  gallon,  and  sold  it  at  $1.20  per  gallon,  on  a  credit  of  3 
months.  While  in  my  possession,  the  rum  lost  10  gallons  by 
leakage :  what  did  I  gain  or  lose,  interest  being  at  6  per  cent.  ? 

Ans.  $70.19  gain, 

8.  A  man  buys  596  gallons  of  wine,  at  6sh.  3d.  per  gallon, 
cash,  and  sells  it  immediately  at  6sk.  9d.  per  gallon,  payable  in 
3  months,  interest  at  6  per  cent,  per  annum:  what  does  he  gain? 

A71S.  £11  11  sh.  8d, 

9.  A  distiller  is  offered  1000  gallons  of  molasses,  at  48  cents, 
cash,  per  gallon,  or  50  cents,  with  2  months  credit,  the  interest 
being  at  8  per  cent. :  which  is  the  most  advantageous  bargain, 
and  by  how  much  ?       Atis.  He  will  gain  $13.6  by  cash  paym-ent. 

"^    ^M4  pieces  of  stuff  be  bought  at  $9.60,  and  5  of  them 


2S$  LESSON  LXXI. 

sold  at  $14.40,  and  4  at  $12  :  at  what  price  must  the  rest  be  dis- 
posed of  to  gain  20  per  cent,  on  the  whole  ? 

Jl.  Sold  10  casks  of  alum,  weighing  gross  33cu't.  2gr.  15/J.,  tare 
15lb.  per  cask :  what  is  the  whole  amount  of  sale,  at  23sk.  Ad, 
per  hundred  weight  ?  Ans.  £31  13.sh.  Q^d, 

12.  What  is  the  cost  of  32  boxes  of  soap,  weighing  31,550 
pounds,  at  8  cents  per  pound,  allowing  4  pounds  per  box  for  draft, 
and  12  per  cent,  for  tare  ?  A7is.  $301,211. 

13.  A  man  buys  4  hogsheads  of  tobacco,  weighing  38cwt.  2qr. 
8lb.  gross,  tare  94  pounds  per  hogsiiead,  at  $9  per  hundred 
"weight,  ready  money,  and  sells  them  at  111^  pence  per  pound, 
allowing  tare  at  14  pounds  per  hundred  weight.  He  is  to  receive 
two-thirds  in  cash ;  and,  for  the  remainder,  a  note  at  90  days. 
His  gain  or  loss  is  required,  supposing  he  discounted  the  note  at 
60  days,  at  6  per  cent,  per  annum.  A7is.  $283.43  projit. 

LESSON  LXXI. 
EXCHANGE  AND  THE  REDUCTION  OF  CURRENCIES. 

1.  Exchange  is  that  mercantile  operation  by  which 
remittances  of  money  are  made  from  one  place  or  coun- 
try to  another. 

It  is  called  exchange  because,  when  the  trade  is  well 
regulated,  the  settlement  of  accounts  is  made  by  means 
of  drafts,  without  any  actual  transmission  of  money. 

Thus:  Ay  of  Richmond,  sells  tobacco  to  B,  of  Liver- 
pool. Now,  if  B  were  to  send,  in  return,  goods  for  a 
like  amount,  it  is  very  clear  that  their  account  would  be 
readily  balanced  on  their  books,  without  transmission  of 
money. 

But  the  intricacies  of  commerce  do  not  allow  of  so 
easy  a  settlement.  Goods  are  imported  by  different  mer- 
chants :  C,  of  Richmond,  for  example,  may  receive  some 
from  another  person,  D,  of  Liverpool.  Then  he  applies 
to  A  for  a  draft  on  B,  of  Liverpool,  and  sends  it  to  Z>, 
who  collects  the  amount  from  his  neighbor,  B.  So  much 
of  the  account  between  the  two  places  is  settled  by  this 
draft ;  and  the  whole  may  be  so  settled  unless  the  balance 
on  either  side  is  considerable ;  in  which  case,  coin  has  to 
be  remitted. 

Sometimes  the  operation   is   more    complicated,  and 


EXCHANGE.  287 

drafts  have  to  pass  through  several  hands :  C,  for  instance, 
may  find  no  other  channel  of  remittance  than  through 
New  York  and  Paris ;  that  is,  he  may  be  compelled  to 
procure,  from  a  merchant  in  New  York,  a  draft  upon  a 
merchant  of  Paris,  which  he  remits  to  D.  The  draft 
then  finds  its  way  from  Liverpool  to  Paris,  in  some  mer- 
cantile operation  between  the  two  places. 

2.  M  an  exact  balance  of  accounts  between  several 
countries  existed  at  all  times,  settlements  by  drafts  would 
always  suffice,  and  nothing  more  would  be  requisite  than 
to  reduce,  in  the  accounts,  the  currency  of  one  country 
to  the  standard  of  the  other,  in  order  to  establish  the 
equality  of  amounts  in  the  books. 

3.  But  exchange  is  affected  by  the  increase  or  diminu- 
tion of  the  bills  drawn  in  one  country  on  the  other, 
which,  like  other  commodities,  rise  or  fall  according  to 
the  demand  ;  that  is,  when,  at  any  particular  time,  funds 
on  a  particular  place  are  in  great  demand,  there  is  an 
advance  on  their  intrinsic  value,  and  the  holder  of  a  draft 
is  allowed  a  'premium^  the  limit  of  which  is  evidently  the 
cost  of  conveying  the  specie  from  the  debtor  to  the  cre- 
ditor. When,  on  the  contrary,  the  indebtedness  is  on 
the  other  side,  drafts  are  readily  obtained ;  they  are  then 
at  a  discount, 

•i.  The  premium  or  discount  for  100,  is  the  rate  of  ex- 
change. 

When  there  is  neither  premium  nor  discount,  but  an 
exact  intrinsic  equivalency  exists  between  the  currencies 
of  two  countries,  exchange  is  said  to  be  at  par, 

5.  The  money  or  currency  of  a  country  has,  therefore, 
two  distinct  values :  its  intrinsic  valve  and  its  commer- 
cial value. 

The  intrinsic  value  is  that  which  a  coin  has,  when 
compared  to  a  certain  standard,  according  to  the  quantity 
of  each  metal  it  contains. 

Tables  of  the  relative  values  of  coins  are  prepared  in 
every  country,  for  the  use  of  commercial  men,  generally 
in  reference  to  the  particular  standard  of  their  own  coun- 


288  LESSON   LXXI. 

try.     I  will  only  give,  at  the  end,  a  table  of  the  coins, 
the  value  of  which  has  been  fixed  by  Congress. 

The  commercial  value  varies  with  the  fluctuations  of 
trade,  and  is  equal  to  the  intrinsic  value,  with  the  addi- 
tion or  subtraction  of  the  exchange,  according  to  the 
current  rate  at  the  time. 

6.  In  order,  therefore,  to  transfer  an  amount  of  money 
from  one  country  to  another,  the  relative  intrinsic  value 
in  both  and  the  rate  of  exchange  must  be  combined  in  a 
compound  proportion.  The  operation  is  one  of  great 
simplicity. 

I.  You  change  the  sum  of  money  into  another  cur- 
rency by  multiplying  it  into  the  intrinsic  ratio  of  the 
two  currencies, 

n.  Then,  you  add  or  subtract  the  amount  of  exchange^ 
according  to  the  current  rate. 

7.  The  first  part  of  the  operation  is  known  as  the 

REDUCTION    OF    CURRENCIES. 

Example  1st, — Let  it  be  proposed  to  estimate  a  sum  of 
£165  9sh,  6d.,  Georgia  currency,  in  Virginia  currency. 

By  reference  to  the  note  in  Lesson  XLVIIL,  we  find 
that  $1  is  equal  to  6sh.  Virginia,  and  4fsh,  Sd,  Georgia 
currency.     Hence  we  have 

4^sh,  8d.  :  £165  9sh.  6d.  :  :  6sh,  :  (£165  9sh.  6d.)  X 

Ash.  Sd. 

That  is,  we  must  multiply  the  given  sum  by  the  ratio, 

j-T'kd  °^"^'  ^^^  *^^  answer  is,  £212  Ibsh.  OfcZ. 

The  same  might  also  be  obtained  in  multiplying  by  the 
ratio  of  the  respective  currencies,  expressed  in  terms  of 
the  pounds,  reduced  to  Federal  money  j  which  is, 

^  =  ^,  as  above  (see  same  note). 

Example  2d. — If  the  same  amount  was  to  be  reduced 


EXCHANGE. 


289 


to  Federal  money,  a  simple  multiplication  by  the  value 
$4f  of  the  Georgia  pound  would  suffice : 

Observing  that  Aj  =  7^,  and  reducing  the  inferior  denomina- 
tions to  decimal  fractions  of  the  pound  (LI.,  4),  we  would  get 

$VX  165-4.75  =$709,178. 

These  operations  are  too  simple  to  need  further  ex- 
amples. 

EXCHANGE. 

8.  Exchange  may  take  place  between  two  countries 
which  use  the  same  currency ;  as,  for  example,  between 
New  York  and  Richmond. 

If,  for  instance,  exchange  were  10  per  cent,  in  favor 
of  New  York ;  that  is,  if  f  1  in  New  York  were  worth 
$1.10  in  Richmond,  and  I  wished  to  place  $1000  in  the 
former  place,  I  must  pay  $1,100  in  Richmond  for  a  draft. 

If,  on  the  contrary,  exchange  was  10  per  cent,  against 
'New  York;  that  is,  if  $1  in  Richmond  were  worth  $1.10 
in  New  York,  New  York  funds  would  be  at  a  discount  of 
10  per  cent.;  and  the  draft  on  New  York,  for  $1000, 
would  cost  only  $909.09,  as  results  from  the  proportion 
1.10  :  1  :  :  1000  :  Yf^  =  909.09. 

This  is  the  correct  mode  of  computation;  but,  gene- 
rally, the  premium  is  simply  deducted  in  a  manner  simi- 
lar to  bank  discount. 

Exchange,  when  the  currencies  are  different,  combines 
the  two  preceding  operations. 

Example  1st, — Exchange  on  London  being  at  7i  per 
cent,  premium,  I  wish  to  remit  £1,120:  how  much  is 
to  be  paid  here  for  a  draft  of  this  amount  ? 

Ans,  $4.44.X  1120x  1.07J-  =  $5,345.76. 

N.  B.— By  the  laws  of  Congress,  the  pound  sterling  was  reck- 
oned at  $4.44;  but,  more  recently,  Congress  having  lowered  the 
standard  of  gold  coin  in  the  United  States,  the  par  or  intrinsic 
value  of  the  pound  sterling  has  been  increased.  It  is  now  about 
$4.85;  yet,  in  computing  exchange,  $4.44  is  retained  for  the 
value  of  the  pound  sterling,  but  a  certain  percentage  is  added  to 
equalize  the  exchange  :  so  that  an  addition  of  9 J  per  cent.,  which 

25  T 


290  LESSON   LXXII. 

raises  S4.44  to  $4.85,  establishes  par;  and  exchange  at  9 J  is 
therefore  considered  equal  to  par. 

Frequently  the  rate  of  exchange  is  given  by  quoting 
the  value  of  the  unit  of  currency  of  the  foreign  country, 
instead  of  a  percentage.  This  is  the  case  in -regard  to 
exchange  upon  France,  as  in  the  following  question : 

Example  2d, — The  exchange  on  France  being  quoted 
dX  francs  5.35,  v^hat  should  be  paid  for  a  draft  of  6,519 
francs  ? 

This  quotation  means  that,  according  to  the  rate  of  ex- 
change, the  dollar  is  worth  5.35  francs.  We  therefore 
divide  the  amount  of  francs,  6,519,  by  5.35, 

Ans,  $1,218.50. 

Example  3d. — A  purchase  has  been  made  in  Russia,  to 
an  amount  of,  2,480  rubles,  by  a  merchant  of  Richmond, 
who  has  no  direct  means  of  making  payment.  He  is 
obliged  to  purchase  in  New  York  a  draft  on  London, 
which  he  remits  to  St.  Petersburg.  The  exchange  oa 
New  York  is  at  a  premium  of  f  per  cent. ;  the  exchange 
on  London  is  at  9|  premium  in  New  York;  and,  finally, 
the  rate  of  exchange  between  St.  Petersburg  and  London 
is,  ruble  2sh.  d^d. :  what  should  be  the  amount  of  the 
draft  in  sterling,  and  how  much  should  be  paid  in  Rich- 
mond ? 

Ist.  Ans.  2480 X  (25^.  9^^.)  =  £346  3sh.  4>d. 

(face  of  the  note.) 

2d.  Ans.  $4.44  X  1 .0075  x  1 .0975  x  (^£346  3sh.  4cZ.) 
=  $1,699,487,  to  be  paid  in  Richmond, 

Questions. — What  is  exchange  ?  How  is  it  made?  What  are 
drafts  ?  What  is  rate  of  exchange  ?  When  is  it  at  a  premium  ? 
At  a  discount  ?  How  many  kinds  of  values  has  any  currency  ? 
What  is  its  intrinsic  value  ?  Its  commercial  value  ?  How  is 
currency  reduced  ?     How,  exchange  calculated  ? 

EXERCISES. 

1.  Change  $237.50  to  English;  New  York,  New  England,  and 
Virginia;  'Pennsylvania,  Georgia  currency. 

2.  Change  6,245  thalers  of  Prussia  to  Federal  money  (see  at 
the  end.) 


COMPOUND   INTEREST.  '291 

3.  Change  4,243  thalers  to  rubles  of  Russia. 

4.  Change  $647.50  to  florins  of  Austria. 

5.  When  exchange  on  London  is  at  SJ,  what  must  be  paid  for  a 
draft  of  X4,670  lOsh.l 

6.  Exchange  being  at  francs  5.44,  what  is  to  be  paid  fot 
10,250.75  francs  ? 

7.  New  York  funds  being  at  a  premium  of  %  in  Richmond,  and 
at  a  discount  of  If  in  New  Orleans,  what  should  be  given  in  Rich- 
mond to  remit  $7,455.75  to  New  Orleans,  in  New  York  funds  ? 

S.  Cincinnati  funds  being  at  a  discount  of  Ij  per  cent,  in  New 
York,  and  New  York  funds  at  a  premium  of  i  in  New  Orleans,  a 
merchant  of  Cincinnati  directs  his  correspondent  in  New  York  to 
pay  $4,560  for  him  in  New  Orleans :  what  must  he  pay  in  Cin- 
cinnati for  it  ? 

LESSON  LXXIL 
COMPOUND  INTEREST. 

Algebraical  formulae  :  s  =  p  (l-^-r)*,  and  i  =  s  — p, 

1.  Compound  interest  is  that  which  arises  from  adding 
unpaid  interest  to  the  principal,  and  taking  interest  on 
the  aggregate  amount. 

The  same  operation  may  be  repeated  succcessively  at 
each  period  when  interest  becomes  due. 

It  will  be  perceived,  therefore,  that  compound  interest 
is  calculated  by  the  ordinary  rules.  The  following  is  an 
example  of  the  process : 

What  are  the  compound  interest  and  the  amount  of 
$500,  for  6  years,  at  8  per  cent,  per  annum  ? 

OPERATION. 

Principal,  1st  year,  .         .         $500. 

Amountof  the  1st  year,  500X1.08=  540     .    principal,  2d  year. 
«  2d   year,  540X1.08=  583.20  do.  3d  year, 

«  3d   year,  583.20  X  1.08  =  629.856        do.  4th  year. 

«  4th  year,  629.856X1.08=  680.244+  do.  5th  year. 

«  5th  year,  680.244X1.08=  734.664+  do.  6th  year. 

"  6th  year,  734.664 X  1.08  =  793.437,  am't  required.' 

2.  The  compound  interest  is  the  difference  between  the 


292 


.Te        LESSON    LXXII. 


last  amount  and  the  original  principal ;  in  this  example, 
$293,437. 

The  simple  interest  for  the  same  period  would  be  only 
$240.  The  increase,  by  compounding  interest,  is  only 
$53,437  J  but;  though  compound  interest  increases  slowly, 
at  first,  it  swelJs  the  capital  rapidly  when  the  computation 
is  carried  to  a  considerable  period. 

3.  Owing  to  this  rapid  accumulation  of  principal,  the 
exacting  of  compound  interest  is  forbidden  by  law. 

4.  The  computation  of  compound  interest  is  very  la- 
borious by  plain  arithmetical  operations,  but  presents 
otherwise  no  great  difficulty.  Algebra  furnishes  simpler 
methods  to  calculate  it.  For  the  convenience,  however, 
of  those  who  are  not  acquainted  with  algebra,  and  the 
use  of  logarithms,  the  following  table  is  introduced  here : 

TABLE, 

Showing  the  amount  of  ONE,  for  any  number  of  years  up  to  30, 
at  the  rates  of  5  and  6  per  cent,  per  annum. 


Years. 

5  per  cent. 

6  per  cent. 

Years. 

5  per  cent. 

6  per  cent. 

1 

1.050000 

1.060000 

16 

2.182874 

2.510351 

2 

1.102500 

1.123600 

17 

2.292018 

2.692772 

3 

1.157625 

1.191016 

18 

2.406619 

2.854339 

4 

1.215506 

1.262476 

19 

2.526950 

3.025599 

5 

1.276281 

1.338225 

20 

2.653297 

3.207135 

6 

1.340095 

1.418519 

21 

2.785962 

3.399563 

7 

1.407100 

1.503630 

22 

2.925260 

3.603537 

8 

1.477455 

1.593848 

23 

3.071523 

3.819749 

9 

1.551328 

1.689478 

24 

3.225099 

4.048934 

10 

1.628894 

1.790847 

25 

3.386354 

4.291870 

11 

1.710339 

1.898298 

26 

3.555672 

4.549382 

12 

1.795856 

2.012196 

27 

3.733456 

4.822345 

13 

1.885649 

2.132928 

28 

3.920129 

5.111686 

14 

1.979931 

2.260903 

29 

4.116135 

5.418387 

15 

2.078928 

2.396558 

30 

4.321942 

5.743491 

The  following  are  added  here,  merely  to  show  the  rapid  in- 
crease of  compound  interest,  for  long  periods  and  higher  rates. 


50 
100 


11.467392 
131.501 


18.420147 
339.302 


200  I   17,292.51   115,125.84 
300  [2,273,982.75  39,062,430.21 


COMPOUND   INTEREST.  293 

This  table  can  easily  be  extended  :  To  find  the  amount  for  any 
nnniher  of  years  greater  than  the  number  in  the  table,  tmdtiply  to- 
gether the  arnounts  for  two  n^tmbers  of  yean,  whose  surn  is  equal 
to  the  period  for  whicli  the  aTtiount  is  required. 

Thus,  50  may  be  obtained  by  multiplying  the  amount  for  25 
years  by  itself,  or  that  for  24  and  26  together,  or  any  other  two 
amounts  whose  periods  together  make  50. 

5.  The  use  of  this  table  is  very  simple  :  In  order  to 
find  the  amount  for  any  sum  of  money  for  a  number 
of  year ^"^^  multiply  the  amount  standing  against  the  i)eriod 
by  the  given  principal ;  for  it  is  evident,  from  the  man- 
ner of  obtaining  the  amount,  that  it  is  proportional  to  the 
principal. 

6.  Questions  like  those  given  in  simple  interest,  may  also 
be  proposed  in  compound  interest ;  such  as  finding  the  rate^ 
time,  &c.  But  they  require  a  knowledge  of  algebra  and 
logarithms,  and  are  too  complicated  for  arithmetic. 

7.  Though  compound  interest  is  not  allowed  by  law, 
it  is  frequently  important  to  calculate  it,  as  a  basis  for 
speculations,  contracts,  bargains,  notes,  investments,  the 
value  of  stocks,  the  cost  and  character  of  constructions, 
especially  public  works,  in  reference  to  their  duration,  &c. 

8.  Although  I  must  refer  to  algebra,  and  recommend 
its  study  for  a  complete  knowledge  of  this  subject,  I  will, 
in  the  following  lessons,  give  a  few  examples  of  its  ap- 
plications to  the  investigation  of  certain  transactions. 

It  is  frequently  desirable  to  know  when  a  capital  will 
double  itself,  at  a  certain  compound  interest.  The  follow- 
ing method  is  simple  and  sufficiently  correct  for  practical 
purposes,  at  the  usual  rates  of  interest : 

The  number  of  years  in  ivhich  a  capital  is  doubled^ 
multiplied  by  the  rate  of  interest,  makes  nearly  72. 
Thus,  at  5  per  cent.,  it  will  double  in  about  14  years ;  at 
6,  in  about  12,  &c. 

Questions. — What  is  compound  interest  ?  How  is  it  computed  ? 
Does  it  raise  the  amount  rapidly?  How  is  the  table  used?  Is 
it  allowed  by  law  ?  What  questions  may  be  proposed  in  com- 
pound interest  ?  Can  they  all  be  solved  by  arithmetic  ?  Though 
compound  interest  cannot  be  exacted,  is  it  useful  to  consider  it  ? 

.      25* 


294  LESSON   LXXIII. 

In  what  cases  ?     How  will  you  ascertain  when  the  capital  will 
be  doubled,  at  compound  interest  ? 

EXERCISES, 

1.  "What  is  the  amount  and  compound  interest  of  $629,  for  7 
years,  at  6  per  cent.?  Ans»  Amount,  $945.78. 

Interest,  $316.78. 

2.  What  is  the  amount  and  compound  interest  of  $1,256,  for 
,  15  years,  at  0.06  per  annum  ?  Ans,  Amount,  $1,754,066. 

Interest,     $498,066. 

3.  What  is  the  amount  of  $12,500,  for  6  years,  at  5  per  cent.  ? 

Ans,   $16,751.20. 

4.  What  is  the  compound  interest  of  X246  145^.  M.,  for  3 
years,  at  6  per  cent.  ?  A7is,  £A1  2sk.  6d, 

5.  What  is  the  compound  interest  of  £760  10,9^.,  for  4  years, 
at  0.06  per  annum  ?  Ans.  £199  12^/a.  2d, 

6.  What  is  the  compound  interest  of  jG370,  for  6  years,  at  4 
per  cent.  ?  A7is.  £98  3sk.  \\d, 

7.  What  is  the  amount  of  $76.75,  for  2f  years,  at  0.03  per 
annum?  Ans.  $83,257. 

8.  What  is  the  amount  of  $217,  for  2  J  years,  at  5  per  cent,  per 
annum,  the  interest  payable  quarterly?  Ans.  $242,669, 

9.  The  population  of  a  country  is  10,000,  and  increases  at  the 
rate  of  0.01  per  year :  what  will  it  be  in  10  years  ? 

Ans,  11,046. 

LESSON  LXXIII. 
COMPOUND  DISCOUNT. 

5  1 

Algebraical  formulas  :  p  =  ,—~ — -         .    d  ==  s  (1  —  — — — -\ 
(l  +  r)<  {l  +  ryJ 

1.  It  has  already  been  said  that,  though  the  law  does 
not  allow  of  compound  interest,  it  is,  nevertheless,  fre- 
quently necessary  to  compute  it,  with  a  view  to  ascertain 
the  propriety  or  relative  advantages  of  various  transac- 
tions. 

The  fundamental  principle  of  all  computations  where 
compound  interest  is  introduced  is,  that  the  holder  of  the 
money  has  the  power,  by  retaining  it,  to  invest  and  re 
invest  successively  its  proceeds,  and  thus  to  compound 


COMPOUND    DISCOUNT.  295 

interest ;  and  that  he  would  not  consult  his  own  advan- 
tage  by  entering  into  any  agreement  which  would  not 
place  him,  at  the  end  of  a  fixed  time,  in  the  same  situa- 
tion as  if  he  had  retained  the  use  of  his  funds. 

Let  us  suppose,  for  instance,  that  some  property  has 
been  sold  for  $18,000,  payable  in  10  years,  and  that  the 
seller  should  wish  to  have  it  discounted,  the  rate  of  inte- 
rest being  8  per  cent. 

The  purchaser  would  naturally  consider  that,  by  the 
rule  of  simple  discount,  he  would  have  to  pay  #10,000 
cash ;  whereas  the  10,000  dollars,  if  retained  in  his 
hands,  would,  by  the  accumulation  of  semi-annual  inte- 
rest for  10  years,  reach  to  an  amount  of    .     $21,911.22. 

He  must,  therefore,  decline  an  arrangement  at  simple 
interest,  by  which  he  would  be  a  loser,  in  the 
end,  to  the  amount  of    ....         $3,911.22. 

By  the  computation  of  compound  discount,  however  j 
that  is,  by  finding  the  principal  which,  at  semi-annual 
compound  interest  {the  usual  period  for  dividends)  would 
produce  the  amount  to  be  paid,  the  purchaser  would 
ascertain  that  the  sum  he  may  pay,  without  either  loss 
or  profit,  for  $18,000,  due  in  10  years,  at  a  rate  of  8  per 
cent.,  is  only $8,214.96 

That  is,  as  much  as  ...         $1,785.04 

less  than  the  principal  resulting  from  the  computation  by 
simple  discoimt.  A  transaction  upon  such  a  basis  is  the 
only  one  which  would  do  justice  to  both  parties. 

2.  Hence,  compound  discount  is  the  only  correct  mode 
of  calculation,  especially  for  long  periods.  As  regards 
short  periods  and  small  sums,  the  difference  is  generally 
too  inconsiderable  to  be  noticed  j  it  would  only  embarrass 
ordinary  transactions. 

3.  Compound  discount  is  calculated  in  a  manner  simi- 
lar  to  simple  discount. 

I.  To  find  the  present  value,  or  principal,  divide  the 
whole  amount  by  the  compound  amount  of  ONE  for  the 
given  time. 

n.  Subtract  the  result  from  the  given  amount ;  the 
difference  will  be  the  compound  discount. 


296  LESSON   LXXIV. 

4.  Questions  like  the  following  belong  to  compound 
discount,  and  show  some  of  its  applications  : 

1.  What  is  the  present  value  of  $30,000,  payable  in  7  years,  at 
6  per  cent,  interest?  A7is.  $19,951,713. 

2.  A  man  proposes  to  buy  land  for  $4,587,  payable  in  7  years: 
to  what  cash  price  is  it  equivalent,  interest  being  at  5  per  cent.  ? 

A?i.s.  $3,259,897. 

3.  How  much  should  be  invested  at  present,  at  4  per  cent,  per 
annum,  to  make  up  £569  6sh,  Sd.,  to  be  paid  in  9  years  ? 

Ans.   £AOQ. 

LESSON  LXXIV. 
ANNUITIES. 

1.  An  annuity  is  a  sum  of  money  payable  at  regular 
periods,  either  for  a  limited  time  or  for  ever. 

2.  The  amovnt  of  an  annuity,  forborne  for  some  time, 
is  the  sum  of  all  the  payments  due,  with  the  addition  of 
the  interest  on  each. 

3.  The  present  value  of  an  annuity  is  that  sum  which, 
being  properly  invested,  will  be  exactly  sufficient  to  pay 
the  annuity. 

4.  Interest  on  annuities  may  be  calculated  at  simple  or 
at  compound  rates, 

ANNUITIES   AT    SIMPLE    INTEREST. 

rt(t  —  \)\  {^  being  the  final  amount; 

Algebraical  formula  :  s  =  a  (t-] ^ — )  <  «  ^h«  annuity;  t  the  time ; 

(  r  the  rate. 

5.  Example, — What  is  the  amount  of  an  annuity  of 
$500,  unpaid  for  5  years,  at  5  per  cent,  per  annum? 


OPERATION. 

1st. 

Amount  of  $500 

,  for4 

years, 

$600 

2d. 

(( 

a 

for  3 

years, 

575 

3d. 

(C 

<c 

for  2 

years, 

550 

4th. 

(( 

<c 

for  1 

year, 

525 

5th. 

otal  amount, 

« 

just  due,       .        • 
•         •         •        • 

500 

T 

.     $2,750 

ANNUITIES.  297 

This  operatioa  is  too  simple  to  require  further  expla- 
nation. 

6.  As  regards  the  present  value  of  an  annuity,  when 
only  simple  interest  is  allowed,  most  writers  on  the  sub- 
ject have  defined  it  to  be :  A  sum  such  that,  if  put  out 
at  interest,  at  the  rate  allowed,  its  amount,  at  the  end  of 
the  time  of  its  duration,  will  be  the  same  with  the  amount 
of  the  annuity. 

This  view  of  the .  question,  though  supported  by  even 
able  mathematicians,  is  undoubtedly  erroneous :  for,  each 
payment  of  the  annuity  does  not  differ  in  any  way  from 
the  payment  of  a  note,  subscribed  beforehand,  for  the 
same  amount ;  and  certainly  no  one  would  think  of  esti- 
mating the  present  value  of  several  notes,  by  computing 
their  amount  to  a  period  beyond  their  maturity,  and  then 
discounting  this  amount  instead  of  the  notes  themselves : 
for,  such  a  calculation  would  evidently  introduce  the  dis- 
count of  the  interest  added,  and  consequently  some  com- 
pound interest. 

Take  this  simple  example  :  Two  notes,  of  $100  each,  are  due 
in  one  and  two  years  :  what  is  their  present  value  ?  It  is  clear 
that,  by  the  above  method,  the  first  note,  at  the  end  of  the  second 
year,  would  amount,  at  6  per  cent,  interest,  to  $106,  and  that,  in 
discounting  this  amount,  we  would  include  also  the  discount  of 
the  interest,  6,  which  is  added  to  the  note. 

7.  The  correct  way  is  evidently  to  discount  each  pay- 
ment separately,  as  would  be  done  for  notes :  for,  it  is 
obvious  that  the  aggregate  present  value  of  all  the  pay- 
ments is  the  sum  of  the  present  value  of  each. 

It  is  but  rarely,  however,  that  such  an  estimate  can  be 
useful,  it  being  erroneous  to  value  annuities  otherwise 
than  with  compound  interest. 

ANNUITIES   AT    COMPOUND    INTEREST. 

Algebraical  formulae  :.==«.  ^^  +  "^''~^     ;,=:  ^  .  ^^  +  ^^'T^ 

r  ^       r         (l-|-r)* 

8.  Computations  relative  to  annuities  at  compound  in- 
terest, are  very  laborious,  by  the  common  rules  of  arith- 


298 


LESSON   LXXIV. 


$607.75 
578.81 
551.25 
525.00 
500.00 

$2,762.81 


metic.  It  is  then  necessary  to  calculate  the  compound 
interest  for  each  year,  separately,  as  in  the  following 
operation,  which  is  applied  to  the  preceding  example : 

1st.  Amount  of  $500,  at  5  per  cent.,  for  4  years, 
2d.  «  «  «  «        for  3  years, 

3d.  «  «  «  «       for  2  years, 

4th.         «  «  «  «       for  1  year, 

5th.        «  «  «  «       just  due, 

Total  amount,     .... 

Algehra  furnishes  means  to  get  as  readily  the  amount  and 
present  value  for  long  as  well  as  for  short  periods. 
^  9.  But,  with  the  tahle  of  Lesson  LXXII.,  these  calcula- 
tions may  he  made  with  tolerable  facility  as  follows :  I. 
To  get  the  total  amount  of  an  annuity ,  find,  in  the  tahle,  the 
compound  amount  of  one  for  the  given  time;  deduct  a  unit 
from  it ;  divide  hy  the  rate,  and  multiply  by  the  annuity, 

II.  To  get  the  present  worth,  find  the  amount  of  the  an- 
nuity, and  divide  it  hy  the  compoimd  amount  of  one  for 
the  time. 

Thus,  in  the  above  example,  the  compound  amount  of 
ONE  for  5  years,  less  1,  being  0.276281;  this,  divided  by 
the  rate  0.05,  and  multiplied  by  500,  gives  2,762.81  for 
the  total  amount  of  the  annuity,  as  we  had  found  it. 

For  the  present  worth,  divide  this  total  by  the  compound 
amount  1.276281,  and  you  get  2,164.81. 

10.  The  present  value  of  an  annuity  which  is  to  con- 
tinue for  ever,  is  evidently  the  principal  whose  annual 
interest  is  equal  to  the  annuity.     (Lesson  LXVII.,  6.) 

Tables  of  the  amounts  of  an  annuity  of  ONE,  and  also 
of  its  present  value,  are  met  with  in  several  arithmetics;  but 
for  the  purposes  of  this  work,  the  table  of  the  compound 
amount  of  ONE  will  be  sufficient,  and  may  be  used  as  just 
explained;  the  tables  of  annuities  being  very  rarely 
wanted  in  practical  transactions. 


EXERCISES. 


1.  What  will  an  annuity  of  $800  amount  to,  at  5  per  cent,  in- 
terest, in  25  years  ?  A7is.  $38,181,679. 


ANNUITIES. 


299 


2.  A  person  wishes  to  purchase  an  annuity  which 'shall  give 
him,  at  6  per  cent.,  an  income  of  $500,  for  10  years  :  how  much 
must  he  pay  for  it  ?  A?is.  $3,680.04. 

The  following  questions  belong  also  to  annuities  and 
compound  discount,  and  are  very  common  applications 
of  them.  The  general  principle  which  must  guide  in 
solving  them,  is  to  compare  the  amounts  produced  at  the 
end  of  the  stipulated  times : 

3.  What  is  the  intrinsic  value  of  government  stock,  paying 
semi-annually  5  per  cent,  interest,  and  redeemable  in  20  years, 
when  common  interest  is  at  6  per  cent.  ? 

The  intrinsic  value  must  produce,  at  the  compound  interest  of 
6  per  cent.,  an  amount  equal  to  the  capital  invested,  together  with 
the  accumulation  of  interest  at  6  per  cent,  on  the  5  per  cent, 
dividends. 

4.  What  is  the  comparative  value  of  stock  bearing  4  per  cent, 
interest,  redeemable  in  15  years,  and  stock  bearing  6  per  cent., 
redeemable  in  25  years,  when  common  interest  is  at  5  per  cent., 
both  dividends  paid  quarterly  ? 

5.  What  annuity  should  be  laid  aside,  to  rebuild  a  structure 
which  has  cost  $25,000,  and  is  estimated  to  last  15  years,  interest 
at  6  per  cent.? 

6.  What  annuity  should  a  man  pay,  who  insures  his  life  for 
$40,000,  interest  being  at  5  per  cent.,  and  his  probability  of  life 
15  years ;  so  that  the  company  may  clear  2  per  cent,  on  the 
amount  they  will  have  then  to  pay  ? 

7.  What  annuity  should  be  paid  to  a  man  for  an  amount  of 
$35,000,  given  up  by  him,  his  probability  of  life  being  12  years, 
and  interest  at  6  per  cent.  ? 

8.  How  much  should  a  man  pay  at  present,  so  as  to  receive 
$1,500  at  the  end  of  every  year,  for  12  years,  the  interest  being 
reckoned  at  7  J  per  cent.  ?  Ans.  $11,602.91. 

9.  What  should  a  man  buy  the  reversion  of  a  lease  of  20 
years,  to  pay  him  $100  a  year,  so  that  he  may  make  8  per  cent, 
per  annum  on  his  money  ?  A7is,  $981,815. 

10.  What  is  the  comparative  advantage,  as  regards  ultimate 
expense,  of  a  wooden  and  of  a  brick  house,  the  wooden  house 
costing  $2,000 ;  the  brick  house  $2,500 ;  and  there  being  in  the 
frame  house  $120  of  exterior  framing,  to  be  renewed  every  20 
years;  $150  of  Weather-boarding,  to  renew  every  8  years;  $60 
of  additional  painting,  to  renew  every  5  years ;  and,  finally,  the 
insurance  of  the  wooden  house  being  1^  per  cent.;  that  of  the 
>^rick  house  |  per  cent.  ? 


300  LESSON   LXXV. 

LESSON  LXXV. 

EQUATION  OF  PAYMENTS. 

1.  The  object  of  this  rule  is  to  find  the  mean  or 
equated  time  of  payment  of  several  sums,  due  at  different 
times;  so  that,  by  paying  the  aggregate  amount  of  all 
the  notes  at  that  time,  there  may  be  no  loss  of  interest  to 
either  party. 

RULE. 

I.  Find,  by  simple  discount^  the  present  worth  of  each 
note, 

IL  Then  find  in  what  time  the  aggregate  principal 
thus  calculated  will,  by  the  addition  of  interest,  become 
equal  to  the  sum  total  of  all  the  notes. 

This  will  be  the  true  equated  time  for  the  payment  of 
the  whole. 

To  illustrate  this  rule  by  a  simple  example,  let  us  sup- 
pose that,  the  rate  of  interest  being  6  per  cent., 

$106  are  due  in  one  year; 
112  "       in  two  years; 

118  "       in  three  years; 

124  "       in  four  years; 

130  "       in  five  years; 

and  that  it  is  proposed  to  reduce  them  all  to  one  note  oi 
the  aggregate  amount,  $590,  payable  at  a  time  such  that 
neither  party  shall  lose  interest  : 

It  is  clear  that  the  present  value  of  each  note,  in  this 
simple  case,  is  $100 ;  and  that,  consequently,  $500  ready 
money  should  cancel  all  obligations. 

Now,  the  question  is,  when  will  $500  become  $590, 
by  the  addition  of  interest  ?  The  interest  being  $90,  all 
we  have  to  do  is  to  find  the  time,  by  Case  III.^  of  Lesson 
LXVII.  J  the  answer  will  be 

90  X  100 


'  EQUATION   OF   PAYMENTS.  301 

When  simple  interest  alone  is  admitted,  this  is  the  cor- 
rect and  equitable  method  to  calculate  the  equated  pay- 
ment, though  it  is  not  the  rule  usually  adopted  among 
merchants,  which  is  as  follows : 

Multiply  the  amount  of  each  note  by  the  time  it  has  to 
run,  and  divide  the  aggregate  of  the  products  by  the 
whole  amount  due;  the  quotient  is  taken  for  the  mean 
time. 

That  this  rule  is  incorrect,  will  readily  appear  by  re- 
ference to  the  preceding  example :  for,  by  multiplying 
each  sum  by  its  time,  we  get 

106  X  1  =  106 
112X2=  224 
118  X  3  =  354 
124  X  4  =  496 
130X5=    650 

1830 

And  1830,  the  sum  of  all  the  products,  divided  by  590, 
gives  3^^-g-  years  for  the  equated  time;  by  which  the 
lender  loses,  on  the  present  value,  $2.57 ;  since  the 
whole  amount,  $590,  discounted  for  3A  years,  gives  only 
$497.43. 

Again,  by  this  difference,  he  would  be  in  possession,  at 
the  end  of  the  second  time,  of  $3.60  less  than  by  the 
other  mode  of  computation  j  since  $3.60  is  the  interest 
on  $590  for  /-g-  of  a  year. 

The  only  recommendation  in  favor  of  the  customary 
rule,  seems  to  be  its  greater  simplicity ;  but  this  consi- 
deration is  hardly  sufficient  to  induce  the  lender  to  sub- 
mit to  a  loss  whicili  may  be  considerable  for  large  amounts 
and  distant  payments. 

2.  Indeed,  neither  of  these  rules  does  justice  to  both 
borrower  and  lender.  The  consideration  of  compound 
interest  is  the  only  mode  by  which  this  end  could  be 
readily  attained  :  for,  it  is  evident  that  the  lender,  having 
the  right  to  insist  upon  payment  at  the  time  each  note 
falls  due,  and  having,  from  that  time,  the  management  of 
his  money,  may,  by  proper  investments,  obtain  the  inte- 
26 


302  LESSON   LXXVi 

Test  of  every  sum  which  comes  into  his  hands,  and  that 
nothing  can  prevent  his  thus  compounding  interest. 

3.  Let  us  now  apply  these  considerations  to  the  same 
example. 

According  to  the  common  rule  for  the  equation  of 
payments,  the  lender  receiving  his  money  in  Sf^- years, 
would,  by  the  accumulation  of  interest,  in  1||  years,  be 
in  possession,  at  the  expiration  of  5  years,  of  §658.708. 

Whereas,  if  he  had  received  his  money  for  each  note, 
when  due,  the  following  result  would  have  been  pro- 
duced, at  the  expiration  of  5  years,  by  the  accumulation 
of  interest ; 


By  the  1st  note,  of  $106, 

the  amount    , 

$133,823 

«         2d      «    of     112, 

<( 

.       133.394 

«          3d      "     of     118, 

a 

132.585 

«          4th    «     of     124, 

te 

131.440 

«          5th    «     of     130, 

C( 

130.000 

Total  amount,    . 

$661,242 

Being  a  difference  of  $2,534  for  the  small  amount  of 
8500. 

4.  These  calculations  show  that,  when  compound  inte- 
rest is  not  allowed,  a  person  who  consents  to  equating 
payments  by  the  common  rule,  may  subject  himself  to 
a  considerable  loss,  on  large  sums. 

The  method  by  discount,  which  has  been  given  at  the 
beginning  of  this  lesson,  is  more  nearly  correct. 

Qucstio7is. — What  is  equation  of  payments  ?  What  is  the 
correct  method,  at  simple  interest  ?  What  method  is  generally 
in  use  ?  Which  of  the  parties  does  the  latter  favor  ?  Is  either 
of  these  methods  mathematically  correct  ?  What  is  the  true  and 
equitable  method  ?     Explain  it. 

EXERCISES. 

1.  A  owes  B  $2000,  whereof  $400  are  to  be  paid  in  3  months; 
$600  in  5  months,  and  the  remainder  in  10  months  :  at  what  time 
may  the  whole  be  paid,  without  injustice  to  either  ? 

Ans,  By  the  tisual  rule^  Into.  3da, 

2.  J£l,200  are  due,  |  in  3  months ;  J  in  6  months,  and  f  in  9 
months :  what  is  the  equated  time  of  payment  for  the  whole  ? 

Ans»  5mo.  1\  da* 


PARTIAL   PAYMENTS.  803 

:  3.  C  owes  to  D  $1,400,  to  be  paid  in  3  months ;  but  D  being 
in  want  of  money,  C  pays  him,  at  the  expiration  of  2  months, 
$1,000:  how  much  longer  than  3  months  ought  C,  in  equity,  to 
defer  the  payment  of  the  rest?  Ans,  2\mo, 

In  the  preceding  examples,  the  times  are  so  short  that  the 
method  by  discount  would  make  but  little  difference.  Let  the 
pupil  verify  it. 

4.  There  fell  due,  on  January  13th,     .        .        .         $700 


on  February  15th, 
on  March  6th, 
on  March  21st, 
on  May  10th, 
on  July  17th, 


1,300 
4,500 
2,200 
5,620 
2,410 


$16,730 

From  what  equated  time  should  interest,  at  6  per  cent.,  be 
charged  on  the  whole  amount,  the  settlement  taking  place  on  the 
1st  of  September  ?  Ans,  By  the  common  ruhy 

By  discount^ 

5.  An  individual  has  bought  an  estate  for  $90,000,  payable  in 
15  equal  annual  instalments,  the  first  of  which  in  12  months ;  the 
parties  agree  to  reduce  the  fifteen  payments  to  one  note  :  at  what 
time  should  it  be  made  payable — 

Ans,  1st,  by  the  common  rule,  .         S  years  ; 

2d,  by  simjjle  discount,  .  7.219; 

^d,  by  compound  interest,        .  7.46. 

What  is  the  loss  or  gain  to  the  purchaser  by  each  of  these 
methods  ? 

;     LESSON  LXXVL 
PARTIAL  PAYMENTS. 

1.  It  frequently  happens  that  partial  payments  are 
made  at  diiferent  times,  on  bonds  and  notes  bearing  inte- 
rest. The  question  then  occurs :  How  is  a  final  settle- 
ment to  be  made  ? 

The  leading  principle  of  such  transactions  is,  I  believe, 
in  every  state,  that  interest  is  not  to  be  compounded ; 
and,  accordingly,  usage  and  the  decisions  of  the  courts 
of  law  have  established  rules  for  the  purpose,  which  vary 
in  different  places.  Whatever  these  rules  may  be,  they 
are  but  simple  applications  of  what  precedes. 

2.  The  rule  most  commonly  in  use,  and  which  has 


304  LESSON  LXXVI. 

especially  been  sanctioned  by  the  courts  of  Massachu- 
setts and  Virginia,  is  as  follows : 

Deduct  each  successive  payment  (if  it  exceed  the  in- 
terest) from  the  amount  of  both  the  'principal  and  inte- 
rest due  at  the  date  of  such  payment^  and  take  the  re- 
mainder for  a  new  principal ;  upon  which  perform  a 
similar  operation  at  the  next  payment  ;  and  so  07i,  until 
the  time  of  final  settlement. 

But,  if  the  partial  payment  he  less  than  the  interest, 
do  not  deduct  it,  but  merely  reserve  it,  to  be  added  to 
subsequent  payments,  until  the  aggregate  sum  paid  in 
exceeds  the  interest  up  to  the  last  payment,  when  the 
whole  is  deducted,  as  above. 

Example, — A  note,  dated  January  1st,  1830,  promises 
to  pay  $10,000  in  6  months,  with  interest  thereon,  from 
date,  at  6  per  cent. 

This  note  is  not  paid  at  maturity,  but  partial  payments 
are  made,  which  are  endorsed  on  the  note,  as  follow : 


Received 

,  April  1st,  1830, 

$240 

t( 

August  1st,  1830,      . 

40 

(C 

December  1st,  1830, 

60 

(C 

February  1st,  1831, 

600 

cc 

July  1st,  1831, 

400 

(, 

June  1st,  1834, 

3,000 

cc 

September  1st,  1834, 

120 

cc 

January  1st,  1835,     , 

150 

cc 

October  1st,  1835, 

500 

Judgment  is  to  be  entered  on  the  1st  of  December, 
1840  :  for  what  amount  should  it  be  rendered? 

According  to  the  preceding  rule,  the  answer  is 
$10,619.62.  The  operations  are  too  simple  to  be  exhi- 
bited here ;  they  are  left  for  an  exercise. 

3.  When  several  notes  are  due,  the  sums  paid  are  ap- 
plied to  the  payment  of  the  first;  and  the  others  are 
considered  in  their  order,  after  the  principal  and  inte^ 
rest  have  been  settled  in  full. 

4.  The  practice  in  Connecticut  differs  from  the  above, 
in  the  following  particulars : 


PARTIAL    PAYMENTS.  §05 

1st.  All  sums  paid,  whether  large  or  small,  are  de- 
ducted from  the  amount  of  principal  and  interest. 

2d.  When  the  remainder  exceeds  the  preceding  prin- 
cipal, interest  is  computed  only  on  this  principal. 

3d.  No  interest  is  reckoned  for  less  than  one  year.  If 
payment  be  made  within  a  less  period,  interest  is  added 
to  both  this  payment  and  the  principal,  to  the  end  of  the 
year,  and  the  difference  between  them  taken. 

This  rule  avoids  compound  interest  more  completely 
than  the  first,  which  is  not  altogether  free  from  it. 

5.  Among  merchants,  in  open  accounts,  interest  is 
frequently  added,  on  both  sides,  to  the  sums  paid  and  re- 
ceived up  to  the  day  of  settlement. 

6.  But,  after  all,  no  rule  which  is  not  based  upon 
compound  interest,  can  be  strictly  correct  or  regular  in 
its  application.  The  above  rule  is  an  example  of  this; 
for,  it  will  be  readily  perceived  that  the  aggregate 
amount  ultimately  received  will  depend  on  the  partial 
payments  made ;  and,  above  all,  on  the  periods  at  which 
they  are  made.  This  should  not  be  the  case ;  and,  what- 
ever be  the  amount  equitably  due,  it  should  be  received 
in  full,  and  the  same  in  every  way  the  payments  may  be 
effected. 

If  compound  interest  were  not  rejected,  this  irregu- 
larity would  not  exist.  The  rejection  of  it  is  mathema- 
tically erroneous,  and  especially  unfavorable  to  the  lenders 
of  money  and  holders  of  notes ;  for,  it  plainly  amounts 
to  preventing  a  man,  who  is  kept  out  of  his  money,  from 
obtaining  the  advantages  which  would  have  accrued  to 
him,  had  he  retained  the  management  of  it. 

7.  Hence,  as  a  natural  consequence,  must  result,  on 
the  part  of  the  debtor,  a  disposition  to  delay  payment, 
and,  on  the  part  of  the  creditor,  less  indulgence  towards 
his  deserving  debtor :  for,  returning  to  the  above  exam- 
ple, it  will  be  easily  ascertained  that  the  person  who  had 
a  right  to  exact  payment  on  the  1st  of  July,  1830,  if  he 
had  received  it  then,  might,  by  successive  reinvestments 
of  the  proceeds,  even  only  every  six  months,  have  in- 

26*  u 


806 


LESSON   LXXVI. 


creased  the  amount,  in  nine  years  and  eleven  months, 

to $17,972.40 

Whereas,  by  the  common  rule,  he  would 
be  allowed,  at  the  end  of  the  time,  if  no 
payments  had  been  made,  only  .         i     $15,950.00 

Surely,  it  will  not  appear  just,  that  the  debtor  should 
be  benefited  to  so  great  an  amount,  while  the  creditor  is 
kept  out  of  his  funds;  and  such  a  result  will  prompt  the 
latter  to  insist  upon  payment. 

If  it  be  considered,  as  the  table  in  Lesson  LXXIL 
shows,  that  compound  interest  does  not  accumulate  rapidly 
at  first,  would  it  not  be  preferable  to  allow  it  for  a  limited 
period ;  ten  years,  for  example  ?  The  policy  of  limita- 
tion might  be  applied,  in  this  case,  with  as  much  pro- 
priety as  in  others. 

Questions, — How  are  partial  payments  generally  settled  ?  Is 
this  rule  just  and  correct  ?  Is  it  favorable  to  the  creditor  or  the 
debtor  ?  What  is  its  natural  tendency  ?  What  is  the  only  w?4y 
by  which  uniform  results  can  be  obtained  ? 

EXERCISE. 
liynchhurg,  Va.,  June  20th5  1841. 

For  value  received,  I  promise  to  pay  Henry  Johnson  or  order, 
on  demand,  three  thousand  eight  hundred  and  forty-nine  dollars 
and  thirty-five  cents,  with  interest  at  6  per  cent,  per  annum. 

John  Hays. 

On  this  note  are  the  following  endorsements : 

March  4th,  1842:  Received  three  hundred  and  seventy-seven 
dollars  and  fifty  cents. 

June  14th,  1843 :  Received  eight  hundred  and  twenty-five 
dollars. 

September  18th,  1843  :  Received  eight  hundred  and  five  dollars. 

June  24th,  1844 :  Received  two  hundred  and  thirty-six  dollars 
and  twenty-five  cents. 

March  Sth,  1845 :  Received  sixty  dollars  and  eighty-five  cents, 

December  9th,  1845  :  Received  four  hundred  and  ninety  dollars. 

July  10th,  1846  :  Received  eight  hundred  and  forty-five  dollars. 
What  remains  due  on  September  28th,  1847  ?      A?is,  $1,131,485. 


CONCLUSION. 

I  HAVE  now  completed  what  constitutes  pure  arithme- 
tic. Indeed,  some  parts  of  what  precedes,  will  be  much 
better  understood  with  the  assistance  of  algebra. 

I  omit  several  subjects  which  do  not  appear  to  belong 
properly  to  arithmetic,  because  it  is  incompetent  to  teach 
them  thoroughly. 

In  the  first  place,  the  rules  of  position^  in  arithmetic, 
are  indefinite  and  complicated ;  whereas,  they  are  of  the 
simplest  order  of  questions  in  algebra,  by  which  they  can 
be  solved  with  great  ease,  after  a  few  lessons  in  this 
science. 

As  regards  involution  and  evolution^  particularly  the 
extraction  of  square  and  cube  roots,  and  also  progressions. 
they  are  applied  chiefly  to  questions  in  geometry  and 
algebra;  and,  therefore,  belong  properly  to  the  latter 
science,  an  elementary  course  of  which  is  the  shortest 
way  to  learn  them,  and,  indeed,  to  obtain  a  thorough 
knowledge  of  arithmetic.  Such  a  course  will  ultimately 
prove  a  saving  of  time. 

The  same  thing  may  be  said  of^wiejmgm^ioy^  which^ 
should  be  preceded  by  the  study  of  geometry,  and  learned 
in  special  treatises.     The  introduction  of  this  subject  in 
arithmetics,  will  rather  delay  than  accelerate  the  progress 
of  the  student. 

For  similar  reasons,  I  omit^ boolcjceeping  and  all  that 
relatestoj^er£an^iZej&2:2a.5i^  There  are  excellent  special 
and  comprehensive  works  on  these  subjects,  which,  for 
study  and  reference,  are  far  preferable  to  the  necessarily 
limited  articles  to  be  found  in  arithmetics. 

To  crowd  so  much  incomplete  and  imperfect  know- 
ledge in  elementary  works,  appears  to  me  calculated  to  do 
more  harm  than  good,  by  deceiving  the  student  into  a 
belief  that  he  has  really  learned  that  which  he  cannot 
well  understand,  and  still  less  apply  practically. 


11 


■\1 


II 


Should  this  small  volume  be  favorably  received,  it  is  my 
tention  to  go  on  with  a  complete  condensed  course  of  math 
matics,  confining  myself  altogether  to  what  is  practically  useful 


he-    ;»iif 


APPENDIX. 


f  OF  COINS. 

i  I  The  purity  of  gold  is  generally  estimated,  not  by  our  weights, 

*  •        but  by  an  Abyssinian  weight,  called  Carat, 

The  carat  is  subdivided  into  4  grai?is,  and  these  into  quarters, 
A  carat  is  equal  to  2^  pennyweights :  pure  gold  is  24  carats 
fine. 

English  standard  gold  is  22  carats  fine,  and  2  carats  alloy, 
which  is  copper. 
f  f       In  the  United  States,  the  standard  of  both  gold  and  silver  has 
I  f    been  sinaplified  by  the  law  of  the  18th  of  January,  1837,  which 

4^     makes  it -^^  of  pure  metal, 

and tV  of  alloy. 

In  gold  coins,  the  alloy  consists  of  pure  silver  and  copper;  the 
silver  never  to  exceed  one  half  of  the  alloy. 

The  weight  of  the  eagle  remains  as  before,  258  grains ;  of 
which  232.2  is  pure  gold. 

That  of  the  dollar,  .         ,         ,         ,        412J  grains ;  of 

which  371.25  is  pure  sjlver. 

And  that  of  the  cent,         .         •         •         168  grains  of  copper. 

This  new  standard  is  also  that  of  France.  So  that  the  compa- 
rative value  of  the  coins  of  the  two  countries  is  now  in  the  exact 
proportion  of  their  weights. 

By  a  subsequent  law  of  March  3d,  1853,  in  consequence  of  the 
great  influx  of  gold  from  California  and  Australia,  and  the  appre- 
ciation of  silver,  the  weight  of  the  fractions  of  the  dollar  were 
reduced  as  follows :  the  half  dollar  from  206^  to  192  grains;  the 
quarter,  dime,  half  dime,  and  three  cent  piece  in  proportion.  No 
silver  dollar  is  now  coined. 

^*  CURRENCY   OF    ENGLAND. 

The  mint  or  standard  price  of  gold  in  England  is,  per  pound, 

troy  weight, £A6  Ush.  Gd. 

,  %^        A  pound  of  standard  silver  contains       lloz.  2dwt.  of  silver, 

and 0      18         of  copper. 

The  proportional  value  of  silver  and  gold  is  as  15^  to  1. 
The  sovereign  of  England  contains,  of  pure 

1      gold,  113.001  grains; 

;         Of  standard  gold, 123.274       " 

f        So  that,  compared  to  the  present  standard  of  the  United  States, 
it  is  intrinsically  worth  •        •        •        •        •        $4.85. 


APPENDIX.  309 

Though,  bv  the  latest  law  of  Congress  on  the  sabject  (March  3d, 
1843),  it  is  made S4.6o9. 

So  that,  to  the  former  value  of  the  pound  sterliog,  $4.44,  fixed 
by  the  law  of  Congress,  of  the  31st  of  July,  17S9,  and  continued 
for  nnany  years,  there  must  now  be  added  9^  per  cent.,  to  give  it 
its  present  value  in  United  States  gold  coin. 

Hence,  exchange  on  England  at  109^  per  cent.,  is  now  consi- 
dered par,  among  mercantile  men,  who  reckon  still  by  the  old 
value,  $4.44,  of  the  pound  sterling. 

The  English  shilling  contains       .        80.727  grains  of  silver, 
and 87.27  of  standard  silver. 

Consequently,  compared  to  the  new  standard  in  the  United 
States,  the  value  of  the  silver  shilling  is       .         .         21|  cents. 

And  the  dollar  is  worth,  intrinsically,  4  silver  shillings  and  l\ 
pence,  though  it  is  generally  reckoned  at  4  shillings  and  6  pence ; 
while  the  shilling,  as  the  twentieth  part  of  a  sovereign,  is  24| 
cents. 

There  is  in  all  this  a  discrepancy,  which  affects  exchange  unfa- 
vorably to  the  United  States ;  the  standard  of  which  is  silver, 
while  that  of  England  is  gold, 

CURRENCY   OF    FRANCE. 

The  standard  of  French  coin  is,  like  that  of  the  United  States, 
0.9  pure  metal  and  0.1  alloy. 

155  gold  pieces,  of  20  francs,  weigh  1  kilogramme,  equal 
to 2.68027tt. 

This  makes  the  single  piece  of  20  francs,  weighing 
6.45161  grammes  =  99.601  grains,  worth  .         .  $3.86. 

One  silver  piece,  of  2  francs,  weighs  1  decagramme  =  154.38355T. 
"  of  5  francs       «'       2.5      «  =  385.9888S5gT. 

Hence,  the  intrinsic  value  of  the  5  franc  piece  is  .  $0.93573, 
and  of  the  United  States  dollar  .         •         .        5.343  francs. 

The  law  of  iMarch  3d,  1$43,  makes  the  20-franc  piece  S3.855, 
and  the  5-franc  piece 0.93. 

It  will  be  remarked  here,  also,  that  4  five-franc  pieces  are  only  S3.74, 
instead  of  the  equivalent  \'alue,  3.86,  of  a  gold  piece  of  2Q  francs :  showing 
that  in  France,  as  well  as  in  England,  the  proportional  rates  of  gold  and  sil- 
ver differ  from  that  adopted  in  the  United  States. 

VALUE    OF  FOREIGN    COINS    MADE    RECEIVABLE  BY  CONGRESS. 

The  Law  of  March  3d,  1843, 
Fixes  the  value  qf  the  gold  coins— 

Of  Great  Britain,  of  not  less  than  0  915|^  in  fineness,  at      80.946  per  dtet. 
Of  France,  of  not  less  than      .      0.S99  "         at        0.929  " 

This  makes  the  Sovereign  of  England  worth         ....        S4.S59 

the  Guinea,  5.102 

and  the  Napoleon,  or  *0  fraiic  piece,  ....  3.S55 

It  also  fixes  the  value  of  the  following  silver  coins : 


glD 


APPENDIX. 


The  Spanish  Pillar  Dollar. 
The  Dollar  of  Mexico,      ' 
The  Dollar  of  Peru,  and 
The  Dollar  of  Bolivia, 


when  not  less  than  I 
■  397  in  fineness,  } 


iwhi 


and    415 

weight, 


grains 


at  100  cents. 


The  Five-franc  Piece,  of  not  less  than  0.9  in  fineness,  and  384  grains  in 
weight,  at  .  .  .  .  .        $0.93 

Milreis  of  Portugal,         .....  1.12 

Rix  Dollar  of  Bremen,  ....  0.78^ 

Thaler  of  Bremen,  of  72  grotes,  .  .  .  0.71 

Milreis  of  Madeira,         .....  l.OO 

Milreis  of  Azores,  .....  0.831- 

Marcbanco  of  Hamburg,  .  .  .  .  0.35 

Ruble  of  Russia,  .....  0.75 

Rupee  of  British  India,  .  .  .  .  0.44^ 

By  the  law  of  May  22d,  1846,  for  computatioii  at  the  custom  house, 

The  dollar  of  Norway  and  Sweden,  ....         $1.08 

The  dollar  of  Denmark,  .....  1.05 

Thaler  of  Prussia  and  northern  States  of  Germany,         .  .  0.69 

Florin  of  southern  States  of  Germany,  .  .  .  0.40 

Florin  of  Austria  and  of  Augsburg,         ....  0.48^ 

Lira  of  Lombardo-Venitian  Kingdom,     ....  0.16 

Franc  of  France  and  Belgiimi ;  Lira  of  Sardinia,  .  .  0.186 

Ducat  of  Naples,     .......  0.80 

Ounce  of  Sicily,       ...*...  2.40 

Pound  of  Nova  Scotia,  N.  Brunswick,  New  Foundland  and  Canada,   4.00 

These  laws  repeal  every  part  of  former  laws,  inconsistent  with 
them ;  and,  consequently,  the  law  of  July  27th,  1842,  whitb  had 
fixed  the  value  of  the  pound  sterling  at  $4.84. 
It  leaves  in  force,  of  the  law  of  June  28th,  1834,  the  rate  of 
The  gold  of  Portugal  and  Brazil,  of  22  carats,  at  $0,948  per  pennyweight. 
"  Spain,  Mexico,  and  Colombia,  of 

20  carats  3  7-16  grains,  .        at      0.899  " 

Calculated  by  these  rates,  the  following  is  the  legal  value  of  the  coins 

Of  Portugal— Dobraon,  of  24,000  Rees,                  .        .  $32,706 

Dobra,  of  12,800  Rees,              .        .        .  17.301 

Johanne.>:,           ...         .        •        .  17.064 

Moidore  (half  in  proportion),          .        .  6.557 

Milree,               0.78 

Piece  of  16  testoons,  or  1,600  rees,          .  2.121 

"     of  12  testoons,  or  1,200  rees,          .  1.574 

"    of    8  testoons,         ....  1.12 

Old  Crusado,  of  400  rees,        .        .        .  0.588 

New  Crusado,  of  480  rees,      .        .        .  0.637 

New  Dobra, 16.253 

Johannes  (double  in  proportion),             .  8.763 

"          (half  in  proportion),               .  4.371 
Of  Brazil — Dobraon,  Dobra,  Johannes,  Moidore,    New  Crusado,  as  in 

Portugal. 
Of  Spain— Quadruple  Pistole  or  Doubloon,  of  1772 ;  double  and  single, 

and  also  shares  in  proportion,           .        .  $16,038 

Doubloon,  1801,            15.535 

Pistole,  1801, 3.884 

Coronilla,  Gold  Dollar,  or  Vintern,  1801,  .983 

Of  Mexico  and  Colombia— Doubloons ;  shares  in  proportion,  15.535 

No  other  coin  is  legal  tender. 

A  great  many  laws  have  been  passed  by  Congress,  in  regard 
to  the  currency,  since  1789;  but  the  above  only  are  now  in  force. 


311 


APPLICATION  OF  THE  RULES  OF  DIVISIBILITY  TO 
PROVE  MULTIPLICATION  AND  DIVISION. 

1.  In  Lesson  XXXVI.,  2,  it  has  been  said  that  a  number  which 
divides  one  factor,  divides  also  the  product. 

Hence,  if  the  multiplicand  or  tmiltiplier  contain  a  certain  fac- 
tor,  verify  if  the  'product  is  divisible  by  the  same.  If  it  is  not, 
you  have  made  some  mistake  in  the  operation.  If  it  is  divisible, 
there  is  much  probability  that  the  operation  is  correct,  though  it 
is  not  a  positive  proof. 

2.  In  Division, 

1.  If  a  factor  exists  in  the  dividend,  and  either  i7i  the  quotient 
or  in  the  divisor,  it  must  be  foitnd  171  the  remaijider. 

XL  If  there  is  a  factor  in  the  divisor  or  in  the  quotient  not  con- 
tained  in  the  dividend,  subtract  the  remainder  from  the  dividend,  and 
the  difference  must  be  divisible  by  this  factor. 

These  are  mere  tests,  but  not  certain  proofs. 

3.  The  most  usual  application  of  the  divisibility  of  numbers, 
however,  is  in  the 

PROOF   OF   MULTIPLICATION   BY   CASTING   OUT   NINES. 

Let  us  suppose  that  we  have  multiplied  2,653,294  by  872,  and 
found  the  product  2,313,672,368.  In  order  to  test  the  correctness 
of  the  operation : 

Find,  by  casting  out  nines,  the  remainder  of  each  factor  divided 
by  9  (XXXIV.,  8);  nnultiply  the  two  rem,ainders  together ;  cast 
the  nines  out  of  this  product ;  the  remainder  of  this  operation 
must  be  the  same  as  that  of  the  given  product^  after  casting  out 
the  nines. 

Thus,  in  the  above  example,  the  remainder 

of  the  multiplicand,  divided  by  9,  is  .  .  .  4 

of  the  multiplier  ......  8 

Their  product  is  32 ;  the  sura  of  whose  figures  is  the  remainder,    5 
which  is  the  same  as  that  of  the  given  product. 

Hence  we  conclude  that  the  operation  is  very  probably  correct. 
But  this  is  not  a  positive  proof;  for,  after  all,  it  merely  esta- 
blishes the  equality  of  the  remainders,  which  may  exist  between 
numbers  entirely  unconnected,  and  may  occur  in  an  incorrect 
multiplication,  by  errors  compensating  each  other.  This  is, 
nevertheless,  a  convenient  test,  and  which  may  be  generally 
relied  on,  with  a  usually  correct  calculator. 


312  APPENDIX. 

Demonstration, — The  reason  of  this  rule  is  obvious ;  for,  both 
the  multiplicand  and  the  multiplier  are  each  composed  of  two 
parts ;  namely :  a  multiple  of  9  and  the  remainder.  Conse- 
quently, in  multiplying  them  together,  the  product  of  the  two 
remainders  is  the  only  part  of  the  general  product  not  divisible 
by  9.  Therefore,  when  nines  are  cast  out  of  it,  there  must  be 
left  the  remainder  of  the  whole  product,  divided  by  9. 

Remark  that,  in  taking  the  sum  of  the  figures  of  a  number,  with  a  view  of 
casting  out  nines,  you  need  not  introduce  any  part  evidently  divisible  by  9. 
(XXXVI.,  6.)  Thus,  in  the  above  multiplicand,  we  sum  up  only  2  +  2,  be- 
cause 6  +  3,  5  +  4,  and  9,  are  each  divisible  by  9;  in  872,  we  reject  72  for 
the  same  reason,  and  get  at  ouce  the  remainder,  8 ;  finally,  in  the  product, 
2,313,672,368,  we  reject  36  j  72  ;  36,  and  sura  up  only  2  +  3  +  1  +  8,  or  even 
2  +  3,  since  1  +  8=9. 

PROOF   OP   DIVISION. 

Division  may  also  be  tested  by  casting  the  nines  out  of  the 
divisor  and  quotient,  and  of  the  dividend,  which  is  their  product. 

If  there  is  a  remainder,  what  is  left  of  it,  after  the  nines  have 
been  cast  out,  must  be  added  to  the  remainder  of  the  product  of 
the  divisor  and  quotient,  and  the  nines  cast  out  of  the  sum,  if 
necessary ;  then  the  final  remainder  must  be  equal  to  that  of  the 
dividend. 


14,716,925,  divided  by  5,375,  gives  a  quotient  2,738,  and  a  remainder  175. 
Now,  the  product  of  the  remainders  of  the  divisor  and  quotient  leaves      4 
and  the  casting  out  of  the  nines  out  of  the  remainder  leaves    ...      4 

which,  added  to  the  above,  gives 8 

Which  is  also  the  remainder  of  the  dividend,  after  casting  out  nines. 

N.  B. — This  test  fails  altogether  in  an  exact  division,  when  the  divisor  is 
divisible  by  9  ;  since,  in  that  case,  any  number  in  the  quotient,  multiplied  by 
the  divisor,  would  give  a  product  divisible  by  9. 


ABBREVIATED  METHOD  OF  APPROXIMATION  IN  THE 
MULTIPLICATION  OF  DECIMALS. 

Decimals  are  most  frequently  used  to  make  calculations  on 
numbers  that  have  been  obtained  by  observations  of  some  kind, 
such  as  measuring,  weighing,  &c. ;  and,  since  we  seldom  can  de- 
pend on  the  accuracy  of  these  observations,  to  within  some 
small  fraction,  it  is  evidently  needless  to  carry  the  calculation 
to  a  degree  of  accuracy  greater  than  the  observations  or  mea- 
surements themselves. 

In  operations  with  decimals,  therefore,  it  is  frequently  unne- 


APPENDIX.  olS 

cessary  to  carry  the  result  beyond  a  limited  number  of  deoma) 
places.  The  operations  of  multiplying  decimals  may  be  much 
shortened  in  such  cases. 

For  example,  to  multiply  2.753  by  2.313,  carrying  the  product 
to  only  three  places  of  decimals,  we  may  proceed  in  three  diiier- 
ent  ways. 


OPERATIONS. 

I. 

II. 

III. 

2.753 
2.313 

2.753 
2.313 

2.753 
2.313 

8259 
2753 
8259 
5506 

i2o 
27 

8 

9 
53 

259 

5506 

826 

28 

8 

6.367689 

6.367 

6S9 

6.368 

The  first  multiplication  is  in  the  usual  way ;  each  succe-ssive 
product  is  shifted  one  step  to  the  left. 

The  second  is  in  reversed  order;  each  partial  product  is 
shifted  one  step  to  the  right. 

In  the  third,  which  is  the  shortened  form,  we  proceed  as  in 
the  second,  only  we  never  write  any  digit  to  the  right  of  the 
column  of  the  last  required  decimal  place ;  we  are  only  careful 
to  add  o7ie  imit  to  the  extreme  digit  when  the  next  appears  to  be 
5  or  upwards.  This  shortened  process  is  particularly  convenient 
when  a  great  number  of  decimals  is  to  be  multiplied. 

N.  B. — Some  arithmeticians  invert  the  figures  of  the  multiplier,  but  the 
method  appears  to  be  more  liable  to  error. 

Decimal  division  is  susceptible  of  a  similar  abbreviation ;  but 
it  requires  a  great  deal  more  care  to  guard  against  mistakes.  On 
this  account,  and  because  its  applications  are  rare,  it  is  unneces- 
sary to  notice  it  here. 


PREPARATORY  TABLES  IN  MULTIPLICATION  AND 
DIVISION. 

In  both  these  operations,  when  the  numbers  are  large,  it  is 
often  convenient  to  make  a  special  table  of  successive  products, 
from  1  to  9.  This  is  particularly  advantageous  in  decimal  divi- 
sions, carried  to  a  great  number  of  places. 

Let  us  suppose,  for  instance,  that  we  have  to  divide 
453,994,781.2346  by  73,809,  to  six  decimal  places  :  it  would  then 
be  convenient  to  prepare  the  following  table  of  products : 

27 


314 


Byl, 

73S09 

2, 

147618 

3, 

2214-27 

4, 

295236 

5, 

369045 

6, 

442854 

7, 

516663 

8, 

590472 

9, 

664281 

10, 

738090 

Every  one  of  which  may- 
be used  as  it  is  wanted, 
as  exhibited  in  the  an- 
nexed operation.  This 
preparatory  arrange- 
ment has  the  addition- 
al advantage,  that  it 
gives  at  once,  without 
trials,  the  exact  figure 
of  the  quotient  for 
each  partial  dividend. 

N.  B. — In  preparing  such 
a  table,  it  is  well  to 
extend  it  to  10,  ar  - 
verification. 


453094781.2346 
442854 


111407 
73809 


3759SS 
3(59045 


694312 
664281 


73609 


6150.940687+ 


507746 
442854 

"648920 
590472 


584480 
516663 


THE  END, 


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